IMAGE  EVALUATION 
TEST  TARGET  (MT-S) 


^  .^. 


V 


:i^ 


^0 


1.0 


1.1 


UilM    |25 
■i'  lii   122 

ISf  lag  ■■ 

^   U£    12.0 


.  "'■     ' 

||l.25  III  1.4      1.6 

^ 

6"     

» 

7] 


/i 


/ 


7 


Photographic 

Sciences 

Corporation 


23  WIST  MAIN  STRUT 

WEBSTiR,N.Y.  14580 

(716)  t73-4503 


'^ 


z 


^ 


^ 


:J 


^^ 


It 


•NJ 


\\ 


.V<^v 


C[HM/ICMH 

Microfiche 

Series. 


CIHM/ICIVIH 
Collection  de 
microfiches. 


Canadian  Institute  for  Historical  IMicroreproductions  /  Institut  Canadian  de  microraproductions  historiquas 


TMhnioal  and  Bibliographic  Notaa/Notaa  tachniquaa  at  bibliographiquaa 


Tha  Inatituta  haa  attamptad  to  obtain  tha  baat 
original  copy  availabla  for  filming.  Faaturaa  of  thia 
copy  which  may  ba  bibiiograpMcaUy  uniqua. 
which  may  altar  any  of  tha  imagaa  in  tha 
raproduction.  or  which  may  algnifieantly  changa 
tha  uauai  mathod  of  filming,  ara  chaelcad  balow. 


□   Colourad  covara/ 
Couvortura  da  coulaur 


r~1  Covara  damagad/ 


Couvartura  ondommagia 


□  Covara  raatorad  and/or  laminatad/ 
Couvartura  raatauria  at/ou  palliculAa 


D 


D 


D 


D 


Covar  titia  miaaing/ 

La  titra  do  couvortura  manqua 


r~]   Colourad  mapa/ 


Cartaa  gtegraphiquaa  wn  coulaur 

Colourad  ink  (l.a.  othar  than  blua  or  blacic)/ 
Encra  da  coulaur  (i.a.  autra  qua  blaua  ou  noira) 


□   Colourad  plataa  and/or  illuatrationa/ 
Planchaa  at/ou  illuatrationa  an  coulaur 

□   Bound  with  othar  matarial/ 
RalM  avac  d'autraa  documarts 


D 


Tight  binding  may  cauaa  thadowa  or  diatortion 
along  intarior  margin/ 

Laraliura  sarria  paut  cauaar  da  I'ombra  ou  da  la 
diatoraion  lo  long  do  la  marga  intArlaura 

Blanic  laavaa  addad  during  raatoration  may 
appaar  within  tha  taxt.  Whanavar  poaaibia,  thaaa 
hava  baan  omittad  from  filming/ 
II  aa  paut  qua  eartainaa  pagaa  bianchaa  ajoutiaa 
lora  d'uno  raatauratlon  apparaiaaant  dana  la  tanta. 
mala,  loraqua  caia  Atait  poaaibia,  caa  pagaa  n'ont 
paa  4t«  filmAaa. 

Additional  commanta:/ 
Commantairaa  supplAmentairaa: 


L'Inatitut  a  microfilmi  la  maillaur  axamplaira 
qu'il  lui  a  *ti  poaaibia  da  —  procurer.  Loa  ditaiia 
da  cat  axamplaira  qui  aont  paut-4tra  uniquaa  du 
point  do  vua  bibliographiqua,  qui  pauvant  modifier 
una  imaga  raproduita,  ou  qui  pauvant  axigar  una 
modification  dana  la  mAthoda  normale  da  fiimaga 
aont  indiquia  ci-daaaoua. 


r~n  Colourad  pagaa/ 


D 


Pagaa  da  coulaur 

Pagaa  damaged/ 
Pagaa  endommagAaa 

Pagaa  raatorad  and/oi 

Pagaa  reatauriaa  at/ou  peliicuMea 

Pagaa  diacolourad.  stained  or  foxei 
Pagaa  dicoioriea,  tacheties  ou  piquiea 

Pagaa  detached/ 
Pagaa  dAtachiaa 

Showthrough> 
Tranaparance 

Quality  of  prin 

Qualiti  InAgale  da  I'impreaaion 

Includaa  supplementary  matarii 
Comprend  du  matiriel  suppl4mentaire 

Only  edition  available/ 
Seule  Mition  disponibie 


□   Pagaa  damaged/ 
Pagaa 

p~]  Pagaa  raatorad  and/or  laminated/ 

rr^  Pagaa  discoloured,  stained  or  foxed/ 

{~n  Pagaa  detached/ 

FT]  Showthrough/ 

r~1  Quality  of  print  variaa/ 

pn  Includaa  supplementary  material/ 

rn  Only  edition  available/ 


Pagaa  wholly  or  partially  obscured  by  errata 
slips,  tissuea.  etc..  hava  been  refilmed  to 
enaure  the  best  possible  image/ 
Lea  pagea  totalement  ou  pertieliement 
obacurciaa  par  un  feuiiiet  d'errata.  una  pelure, 
etc..  ont  iti  fiimiea  i  nouveau  da  fapon  A 
obtenir  la  meiileure  image  possible. 


The  CO 
to  tha 


Tha  in 
poaalb 
of  tha 
fiimini 


Origini 
baglnr 
tholai 
aion,  c 
others 
firatpi 
aion,  I 
or  iliui 


Thai* 
ahall  G 
TINUE 
which 

Mapa, 
diffari 
entirel 
begini 
right  I 
requir 
methc 


Thia  item  is  filmed  at  the  reduction  ratio  chaclced  below/ 

Co  document  est  filmi  au  taux  da  rMuction  indiquA  ci-daaaoua. 


10X 

14X 

18X 

22X 

26X 

»X 

• 

y 

12X 


16X 


20X 


a4x 


28X 


32X 


d«tajit 
ilM  du 
modifisr 
ler  untt 
filmag* 


The  copy  film«d  her*  hat  bMn  raproducMl  thanks 
to  tho  gonorosity  of: 

Douglas  Librsry 
Qussn's  Univsrsity 

Ths  imagss  appaaring  haia  ara  tha  bast  quality 
posslbia  considaring  tha  condition  and  iajiiblllty 
of  tha  original  copy  and  in  icaaping  with  tha 
filming  contract  spacif ieations. 


Original  capias  in  printad  papar  covers  ara  fiimad 
baginning  with  tlia  front  cover  and  anding  on 
tha  last  paga  with  a  printad  or  iliuatratad  impraa- 
sion,  or  ths  back  covar  whan  appropriate.  Ail 
otliar  original  capias  ara  fiimad  baginning  on  tha 
first  paga  with  a  printad  or  iliuatratad  impras- 
sion,  and  anding  on  tlia  kist  paga  with  a  printad 
or  illustrstsd  imprassion. 


iSss 


L'axamptoira  fiimA  fut  raproduit  grica  i  la 
gAnArositA  da: 

Douglas  Library 
Quaan's  University 

l4M  imagae  suivantee  ont  «ti  reproduites  svsc  is 
plue  grand  aoin,  compte  tenu  de  ki  condition  et 
de  hi  nettet*  de  I'exempleire  filmt,  et  en 
conformit*  avac  las  conditions  du  contrat  de 
fiimaga. 

Lae  exempkiires  origineux  dont  le  couverture  en 
papier  eet  Imprimto  eont  filmta  en  commenpant 
par  la  premier  phrt  et  en  terminent  soit  per  le 
demlAre  paga  qui  comporte  une  empreinte 
d'impression  ou  d'illustration,  sdt  par  le  second 
plat,  aelon  le  ces.  Tous  Iss  autras  exemplairas 
origineux  sont  filmte  en  commenpent  per  le 
premiere  pege  qui  comporte  une  empreinte 
d'impression  ou  d'iliustrstion  et  en  terminent  par 
ki  darnlAre  pege  qui  comporte  une  telle 
empreinte. 


The  lest  recorded  freme  on  eech  microfiche 
sImII  contain  the  symbol  •>-<►  (meening  "CON- 
TINUED"),  or  the  symbol  y  (meening  "END"), 
whichever  epplies. 


Un  dee  symboles  suivents  sppersftrs  sur  Is 
dernlAre  imege  de  cheque  microfiche,  ssion  is 
ces:  le  symbols  -^  signifis  "A  SUIVRE",  Is 
symbols  ▼  signifis  "FIN". 


re 


Meps,  pistes,  cherts,  etc.,  mey  be  filmed  et 
different  reduction  retios.  Those  too  large  to  be 
entirely  included  in  one  exposure  ere  filmed 
beginning  in  the  upper  left  hend  corner,  left  to 
right  end  top  to  bottom,  es  meny  frames  as 
rsquirsd.  The  following  diegrems  illustrate  the 
method: 


Lee  certes,  pisnchee,  tableeux,  etc.,  peuvent  Atre 
filmis  i  dee  tsux  de  rMuction  diffArents. 
Lorsqus  le  document  est  trop  grsnd  pour  Atre 
reproduit  en  un  seul  clichA,  11  est  filmA  A  psrtir 
de  I'engle  supArisur  geuche,  de  geuche  A  droite, 
et  de  lieut  en  bee,  sn  prsnent  le  nombre 
d'imeges  nAcsssaire.  Las  diagremmes  suivants 
illuatrent  le  mAthode. 


ly  errata 
Bd  to 

nt 

na  paiure, 

i^on  A 


1  2  3 


32X 


1 

2 

3 

4 

5 

6 

r- 

"''■'■ '  S'.'  ,           "-    '   :    - 

'■'-:■                                                                                                 ■*'..■ 

^-,11 

:V''\ 

l' 

i     I 

4 

yy. 

"':■  .^*. 

* 

i' 

t 

« 

1 

i 

V 

4 

•> 

\ 

r 

1 

THE  ARITHMETIC  OF  CHEMISTRY 


1 


w 


•rl^^o 


\K 


THE   ARITHMETIC   OF 
CHEMISTRY 


BEING 


A   SIMPLE   TREATMENT  OF  THE   SUBJECT 
OF  CHEMICAL  CALCULATIONS 


BY 


JOHN   WADDELL 

B.Sc.  (LoND.),  PH.D.  (Heidelberg),  D.Sc.  (Edin.) 

FORMERLY  ASSISTANT  TO  THE   PROFESSOR  OF  CHEMISTRY 
IN  EDINBURGH  UNIVERSITY 


Wefo  gorft 
THE    MACMILLAN   COMPANY    \ 

LONDON  :   MACMILLAN  &  CO.,  Ltd.  1 

1899 

^//  rights  reserved 


qj34'X\^!7' 


Copyright,  1899, 
By  the  MACMILLAN  COMPANY. 


Kotinooti  i^ress 

J.  S.  Cuahing  &  Co.  —  Berwick  $c  Smith 
Norwood  Mmi.  U.S.A. 


qV 


.-  /  a  ox 


PREFACE 


The  form  that  this  book  has  taken  is  due  to 
a  very  considerable  experience  of  the  difficulties 
encountered  by  students  when  attempting  to  make 
chemical  calculations.  The  effort  has  been  made 
to  smooth  away  these  difficulties,  and  to  give  an 
accurate,  as  well  as  a  simple  and  systematic,  treat- 
ment of  the  subject.  A  number  of  points  are,  I 
venture  to  think,  more  clearly  presented  than  in 
other  books;  for  instance,  Charles's  Law  and  the 
so-called  gramme-molecular  volume  of  gases. 

The  object  of  the  book  is  to  give  clear  ideas 
on  the  subject,  but  a  book  that  serves  this  pur- 
pose must  necessarily  give  the  best  help  to  the 
student  working  for  examinations  that  include  chemi- 
cal calculations.  Hence  many  examples  are  given 
of  questions  set  in  examination  papers  of  a  num- 
ber of  important  universities  in  Britain  and  Amer- 
ica.      These    questions     have    the    advantage    of 


IbO'^ 


48853 


VI 


PREFACE 


presenting  problems  in  different  ways,  so  that  the 
student  is  trained  in  interpreting  different  forms 
of  expression.  It  is  hoped  that  the  book  may  be 
of  service  to  teachers  who  find  difficulty  in  making 
the  subject  clear,  probably  because  they  expect  the 
pupil  to  understand  intuitively  things  which  they 
will  find  he  does  not  understand  unless  stated  in 
the  simplest  form. 

The  attempt  has  been  made  to  make  the  text 
a  continuous  line  of  argument,  and  therefore  some 
matters  that  are  frequently  inserted  as  an  integral 
part  of  a  book  on  calculations  have  been  thrown 
into  an  appendix. 

A  collection  of  equations  in  common  use  appeared 
advisable,  and  the  plan  of  designating  gases  in  some 
way  is,  I  think,  a  good  one. 

I  wish  to  thank  Professor  H.  L.  Wells  of  the 
Sheffield  Scientific  School  of  Yale  University  for 
the  kindly  interest  that  he  has  taken  in  this  little 
volume  and  for  his  valuable  suggestions,  particu- 
larly in  regard  to  the  chapter  on  volumetric  analysis. 

J.  W. 

New  York, 

February,  1899. 


CONTENTS 


in  some 


CHAPTER  I 
Calculation  Weights  of  Elements 

CHAPTER   n 
Simple  Calculations  of  Weights 


•  •  •  • 


PAGE 
I 


CHAPTER   III 
More  Complex  Calculations  of  Weights 


CHAPTER   IV 


The  Volume  of  Gases 


CHAPTER  V 
Calculations  involving  Weight  and  Volume 

CHAPTER  VI 
Calculations  in  Volumetric  Analysis   . 


CHAPTER  VII 


Calculation  of  Formula; 


•  • 


vu 


i8 


•  • 


28 


•         • 


44 


69 


89 


vni 


CONTENTS 


APPENDICES 

A.  The  French  System  of  Measures 

6.  Arithmetical  Calculations    . 

C.  Comparison  of  Thermometric  Scales  . 

D.  Table  of  Calculation  or  Atomic  Weights 

E.  Equations  in  Frequent  Use 

F.  Pressure  of  Aqueous  Vapour 

G.  Table  of  Logarithms  .... 


PACK 
121 

122 

123 

124 

125 

132 


PACE 
121 

122 

124 
125 

132 


ARITHMETIC    OF   CHEMISTRY 


CHAPTER   I 

CALCULATION   WEIGHTS  OF  ELEMENTS 

When  potassium  chlorate  is  heated  strongly  enough, 
oxygen  is  given  off  and  potassium  chloride  is  left 
behind.  The  weight  of  the  oxygen  given  off  is 
39.18%  of  the  original  chlorate,  and  the  potassium 
chloride  is  60.82%.  For  every  100  ounces  or  pounds 
or  grammes  1  of  chlorate  taken,  39.18  ounces  or 
pounds  or  grammes  of  oxygen  are  evolved,  and  60.82 
ounces  or  pounds  or  grammes  of  potassium  chloride 
remain  behind. 

These  numbers  give  the  ratio  that  exists  between 
the  quantities  of  the  different  substances  in  every 
case,  and  so  if  we  wish  to  know  how  much  oxygen 
can  be  obtained  from  18  ounces  or  pounds  or 
grammes  of   potassium  chlorate,  we  require  merely 


1  For  the  system  of  French  measures  see  Appendix  A. 
B  I 


.5 


r  ,-; 


ARITHMETIC  OF  CHEMISTRY 


to  multiply  i8  by 


3918 
100 


and  obtain  as  answer  7.053 


ounces  or  pounds  or  grammes.     The  potassium,  chlo- 
ride left  behind  is,  naturally, 

18  —  7.053  =  10.947  ounces  or  pounds  or  grammes. 

In  the  same  way,  when  pure  marble,  which  is 
calcium  carbonate,  is  heated,  carbon  dioxide  is  given 
off,  and  lime  is  left  behind;  44%  of  the  marble  is 
carbon  dioxide  and  56%  is  lime.  If  we  wish  to 
know  how  muc)i  lime  will  be  left  if  65  ounces  or 
pounds  or  grammes  are  heated,  we  require  merely 
to  multiply  65  by  -^^  and  obtain  as  answer  36.4 
ounces  or  pounds  or  grammec. 

Conversely,  if  the  problem  be  to  determine  how 
much  potassium  chiorate  is  required  to  produce  48 
ounces  or  pounds  or  grammes  of  oxygen,  it  is  evi- 


dent that  we  must  multiply  48  by 


100 
39.18 


,  because 


we  require  100  ounces  or  pounds  or  grammes  of 
chlorate  for  every  39. 1 8  ounces  or  pounds  or  grammes 
of  oxygen. 


48  X 


100 
39.18 


=  122.5  ounces  or  pounds  or  grammes. 


Likewise,  if  the   problem  be   to  determine  how 


CALCULATION  WEIGHTS  OF  ELEMENTS         3 

much  marble  is  required  to  produce  35  ounces  or 
pounds  or  grammes  of  carbon  dioxide,  it  is  evident 
that  35  X  "Y^  =  79-54  ounces  or  pounds  or  grammes 
is  the  answer. 

Water  consists  of  11.11%  hydrogen  and  88.89% 
oxygen,  and  36  ounces  or  pounds  or  grammes  of 
water  contain 


36  X 


and 


36  X 


II. II 
100 


88.89 
100 


=  4  ounces  or  pounds  or  grammes  of 
hydrogen. 


=  32   ounces  or   pounds   or  grammes 


of  oxygen. 

Carbon  dioxide  is  composed  of  27.27%  carbon  and 
73.72%  oxygen,  therefore  to  produce  60  ounces  or 
pounds  or  grammes  of  carbon  dioxide, 

72  1% 
60  X  =  43.64  ounces  or  pounds  or  grammes 

of  oxygen  are  required. 

Now  analysis  and  synthesis  of  substances  give  us 
merely  such  data  as  have  been  presented  above,  but 
chemical  problems  are  rarely  worked  out  in  this  form. 
The  reason  is  this  : 

In  order  to  work  a  problem  in  any  case,  it  would 
be  necessary  to  know  the  percentage  composition  of 


iii 


4  ARITHMETIC  OF  CHEMISTRY 

the  substance  in  question,  and  if  one  were  called 
upon  to  remember  the  percentage  composition  of 
even  the  ordinary  substances,  the  memory  would  be 
severely  taxed.  It  is  found  that  by  assigning  a  defi- 
nite value  to  each  element,  calculations  involve  less 
tax  upon  the  memory,  and  moreover  provide  for  a 
reasonable  theory  as  to  the  structure  of  matter.  For 
the  purpose  of  calculations,  however,  the  theoretical 
development  may  be  left  out  of  sight. 

In  the  percentage  composition  of  water  given 
above,  it  will  be  seen  that  the  weight  of  oxygen  is 
8  times  as  much  as  that  of  the  hydrogen,  or  \  of 
the  total  weight  of  water ;  and  the  problem  to  deter- 
mine how  many  ounces  or  pounds  or  grammes  of 
oxygen  are  in  36  ounces  or  pounds  or  grammes 
of  water,  is  solved  by  multiplying  36  by  |. 

36  X  I  =  32   ounces   or  pounds   or  grammes  of 
oxygen. 


!  I  ' 


Carbon  dioxide,  as  we  saw,  contains  72.73%  of 
oxygen;  carbon  monoxide  contains  57.14%.  Since 
the  monoxide  contains  less  oxygen  than  the  dioxide, 
let  us  take  the  number  8  (which  we  used  in  the 
case  of  water)  to  represent  the  weight  of  oxygen  in 


CALCULATION  WEIGHTS  OF  ELEMENTS 


5 


carbon  monoxide,  and  let  us  now  calculate  how  much 

carbon  is  united  to  8  ounces  or  pounds  or  grammes 

of  oxygen  in  carbon  monoxide. 

Carbon  monoxide  contains  57.14%  of  oxygen  and 

42.86 


42.86%  of  carbon,  therefore  the  carbon  weighs 


57.14 


as  much  as  the  oxygen ;   and  as  the  weight  of  the 
oxygen   was  8,  the  weight  of  the  carbon  must  be 

42.86 


8x 


57.14 


=  6; 


therefore  6  ounces  or  pounds  or  grammes  of  carbon 
are  united  with  8  ounces  or  pounds  or  grammes 
of  oxygen  in  carbon  monoxide.  Carbon  monoxide 
burns  in  oxygen,  to  form  carbon  dioxide,  and  so  if 
we  start  with  6  ounces  or  pounds  or  grammes  of 
carbon,  it  can  be  first  converted  into  carbon  mon- 
oxide, and  then  still  farther  into  dioxide.  But  the 
carbon  in  carbon  dioxide  is  27.27%,  the  oxygen 
being  72.73%.  The  problem  now  is  to  determine 
how  much  oxygen  is  united  with  6  ounces  or  pounds 
or  grammes  of  carbon  in  carbon  dioxide.  For  this 
we  have 


6x 


72.73 
27.27 


=  16   ounces  or  pounds    or    grammes    of 
oxygen. 


ARITHMETIC  OF  CHEMISTRY 


i; 


11 


For  6  parts  by  weight  of  carbon,  there  is  in  car- 
bon monoxide  8  parts  by  weight  of  oxygen,  and  in 
carbon  dioxide  i6  parts  by  weight  of  oxygen,  or 
twice  as  much  in  the  second  case  as  in  the  first. 
In  fact,  this  is  the  meaning  of  the  names  given 
to  the  two  gases,  and,  till  this  relationship  was  dis- 
covered, the  names  could  not  have  been  given.  (The 
gases  were  called  originally  by  other  names  which 
did  not  show  their  relationship.) 

Carbon  is  combined  with  hydrogen  in  marsh  gas; 
75%  being  carbon  and  25%  hydrogen.  For  every 
6  ounces  or  pounds  or  grammes  of  carbon,  there 
are,  therefore,  2  ounces  or  pounds  or  grammes  of 
hydrogen.  In  acetylene  (the  gas  which  has  lately 
come  into  prominence  as  an  illuminant)  92.3%  of 
carbon  is  united  with  7.*]%  of  hydrogen,  that  is, 
6  ounces  or  pounds  or  grammes  of  carbon  is  united 
with  \  ounce  or  pound  or  gramme  of  hydrogen, 
or  twelve  parts  by  weight  of  carbon,  with  one  part 
by  weight  of  hydrogen.  In  all  of  these  calcula- 
tions, we  have  found  that  the  numbers  to  be  used 
for  hydrogen,  oxygen,  and  carbon,  respectively,  are 
I,  8,  and  6,  or  multiples  of  these  numbers.  If  the 
ordinary  symbols  be  used  to  represent  the  elements. 


:i 


CALCULATION  WEIGHTS  OF  ELEMENTS 


then  9  ounces  or  pounds  or  grammes  of  water 
might  be  represented  by  the  formula  HO ;  14 
ounces  or  pounds  or  grammes  of  carbon  monoxide 
by  the  formula  CO ;  and  22  ounces  or  pounds  or 
grammes  of  carbon  dioxide  by  the  formula  COg. 
But  it  is  found  that  while  by  electrolysis  of  water 
all  of  the  hydrogen  that  is  in  the  water  is  set  free 
as  a  gas,  and  \  of  the  weight  of  water  decomposed 
is  hydrogen ;  on  the  other  hand,  when  sodium  acts 
on  water,  only  one-half  as  much  hydrogen  is  set 
free,  that  is,  -^-^  of  the  weight  of  the  water  acted 
upon.  For  this  and  a  number  of  other  reasons,  it 
is  considered  better  to  take  as  our  standard  weight 
of  water,  18  ounces  or  pounds  or  grammes,  of  which 
2  are  hydrogen  and  16  are  oxygen,  and  this  quantity 
is  represented  by  the  formula  HgO,  and  16  ounces 
or  pounds  or  grammes  represent  the  standard  weight 
of  oxygen.  In  the  same  way,  the  standard  weight 
of  carbon  is  12,  and  the  formula  CO  represents 
28  ounces  or  pounds  or  grammes  of  carbon  mon- 
oxide, and  CO2  represents  44  ounces  or  pounds  or 
grammes  of  carbon  dioxide. 

Just  as  there  is  a  standard   number  for  hydro- 
gen, oxygen,  and  carbon,  so  there  is  for  potassium, 


8 


ARITHMETIC  OF  CHEMISTRY 


chlorine,  iron,  zinc,  and  all  the  elements.  These 
numbers  are  usually  called  atomic  weights  and  are 
considered  as  connected  with  the  atomic  theory,  but 
if  the  whole  atomic  theory  were  proved  unfounded, 
these  standard  numbers  could  still  be  used  for  cal- 
culations, and  we  may  call  them  calculation  weights. 
One-half  of  the  difficulty  experienced  by  the  begin- 
ner in  making  calculations  is  that  he  uses  his  sym- 
bols and  formulae  in  a  loose,  slipshod,  and  unscientific 
manner. 

It  is  inaccurate  to  say  HgO  is  a  colourless,  taste- 
less liquid ;  COg  is  a  poisonous  gas  ;  NHg  has  a  pun- 
gent odour.  HgO  represents  i8  parts  by  weight, 
be  it  ounces  or  pounds  or  tons  or  grammes  or  kilo- 
grammes of  water.  CO2  represents  44  parts  by 
weight  of  carbon  dioxide,  and  NHg  represents  17 
parts  by  weight  of  ammonia,  and  the  symbols  should 
never  be  used  instead  of  the  name  of  the  substance 
(by  any  one,  at  all  events,  that  has  the  slightest 
difficulty  in  chemical  calculations). 


tits.  These 
;hts  and  are 
:  theory,  but 
[  unfounded, 
used  for  cal- 
tion  weights. 
)y  the  begin- 
ises  his  sym- 
d  unscientific 

ourless,  taste- 
[^3  has  a  pun- 
s  by  weight, 
nmes  or  kilo- 
44  parts  by 
epresents  \^ 
mbols  should 
:he  substance 
the   slightest 


CHAPTER  II 

SIMPLE  CALCULATIONS  OF  WEIGHTS 

In  making  chemical  calculations,  it  is  necessary  to 
know  the  formulae  employed  for  the  standard  weights 
of  the  substances  which  enter  into  the  reaction.  It 
is  also  necessary  to  know  the  calculation  weights  of 
each  of  the  elements  composing  the  substances,  in 
order  to  determine  the  standard  weights  of  the  sub- 
stances themselves.  It  is,  thirdly,  necessary  to  know 
the  nature  of  the  chemical  reaction  which  takes  place 
and  to  be  able  to  express  it  quantitively  as  is  done  by 
a  chemical  equation.  For  instance,  if  we  are  to  cal- 
culate how  much  oxygen  can  be  obtained  from  a 
given  quantity  of  potassium  chlorate,  we  must  know 
the  formula  of  the  standard  weight  of  potassium 
chlorate ;  namely,  KCIO3.  For  our  purpose  now,  it 
is  unimportant  to  know  zvhy  the  formula  KCIO3  is 
given  and  not  KgClgOg,  which  would  have  exactly  the 
same  percentage  composition.  We  must,  secondly ^ 
know  the  calculation  weights  of  K,  CI,  and  O.     We 

9 


i         ! 


I 


I         i 


10 


ARITHMETIC  OF  CHEMISTRY 


thirdly  require  the  equation  which  represents  the 
action  that  takes  place  when  potassium  chlorate  is 
heated;  namely, 

2KCIO8     =     2KCl-f3  02^ 

The  calculation  weight  of  K  is  39,  of  CI  is  35.5, 
and  of  O  is  16. 

The  equation  then  represents 


2  KCIO3     = 

2(39  +  35-5  +  48) 

\ 
or,  expressed  in  words. 


2  KCl     +     3  O2 

2(39  +  35-5)      3  X  32 


2  X  122.5  =  245  ounces  or  pounds  or  grammes  of 
potassium  chlorate  yield 


and 


2  X  74.5  =  149  ounces  or  pounds  or  grammes  of 
potassium  chloride. 


3  X  32  =  96  ounces  or  pounds  or  grammes  of 
oxygen. 


1  We  are  not  yet  in  a  position  to  discuss  why  this  equation  is  better 

than 

KCIO3     =     KCl  +  3  O 


or 


KCIO3     =     KCl  +  O3 


Either  of  these  would  be  sufficient  for  the  calculation  we  have  at  pres- 
ent in  hand,  but  in  order  to  avoid  having  to  revise  the  form  of  the 
equation  later  on,  it  is  given  as  above. 


SIMPLE  CALCULATIONS  OF  WEIGHTS 


II 


Ition  is  better 


If  now  we  are  asked  to  calculate  how  much  oxygen 
can  be  obtained  from  i8  ounces  or  pounds  or  grammes 
of  potassium  chlorate,  we  have 

1 8  X  -^  =  7-053  ounces  or  pounds  or  grammes  of 


245 


oxygen, 


the  same  result  as  was  obtained  before  (page  2). 

When  hydrogen  is  obtained  by  the  action  of  zinc 
on  sulphuric  acid,  we  have  the  equation 

Zn  4-  H2SO4     =     ZnS04  +  Hg 

It  should  never  be  forgotten  by  the  student  that 
equations  are  not  arrived  at  in  a  purely  algebraic 
fashion.  They  merely  represent,  in  symbols,  quanti- 
tative relationships  that  have  been  discovered  by 
experiment.  For  example,  the  above  relationship 
holds  only  if  the  sulphuric  acid  be  dilute.  If  it  be 
strong,  other  substances  besides  hydrogen  are  formed, 
such  as  hydrogen  sulphide,  sulphur  dioxide,  and  sul- 
phur. An  equation  rarely  represents  the  entire 
chemical  reaction,  but  it  represents  the  chief  one 
under  certain  conditions,  and  in  carrying  out  the 
experiment  the  endeavour  is  made  to  secure  the 
proper  conditions.     The  water  that  is  used  to  dilute 


12 


ARITHMETIC  OF  CHEMISTRY 


the  sulphuric  acid  is  not  mentioned  in  the  equation, 
because  its  amount  is  unchanged  by  the  reaction. 
The  equation,  coupled  with  the  calculation  weights 
of  the  different  elements,  provides  the  data  necessary 
for  the  solving  of  problems. 
The  equation  is,  in  this  case, 


Zn     +     H2SO4     =     ZnS04     + 

65.5         (2  +  32  +  64)  (65.5  +  32  +  64) 
98  161.5 


H. 


That  is,  65.5  grammes^  of  zinc  acting  on  98  grammes 
of  sulphuric  acid  produces  16 1.5  grammes  of  zinc 
sulphate  and  2  grammes  of  hydrogen.  From  the 
information  provided  by  the  equation,  it  is  possible 
to  make  any  calculations  which  involve  merely  the 
weights  of  these  four  substances;  for  instance,  how 
much  zinc  is  required  in  the  preparation  of  10 
grammes  of  hydrogen,  or  how  much  zinc  sulphate 
would  be  obtained  by  the  action  of  150  grammes 
of  sulphuric  acid  on  zinc,  and  how  much  zinc 
would  be  left  unacted  on,  if  200  grammes  of  zinc 
had  been  used  in  the  last  case } 

^  The  English  weights,  such  as  ounces  and  pounds,  are  unusual 
in  scientific  chemistry,  and  are  therefore  omitted  in  the  sequel. 


C 


tion, 
:tion. 
ights 
ssary 


immes 
)f  zinc 
m  the 
ossible 
y  the 
e,  how 
of    10 
Iphate 
ammes 
1    zinc 
of  zinc 


;  unusual 


SIMPLE  CALCULATIONS  OF  WEIGHTS         1 3 

In  the  preparation  of  carbon  dioxide,  marble  is 
acted  on  by  hydrochloric  acid,  and  the  equation  is 

CaCOg    +    2  HCl   =    CaClg   4-   COj   4-    H2O 

(40+12  +  48)     2(1  +  35.5)       (40 +70      (»2  +  32)      (2+16) 
100  73  III  44  18 

If  the  question  is  asked,  how  much  hydrochloric  acid 
is  required  for  the  preparation  of  60  grammes  of 
carbon  dioxide,  we  have 

60  X  IJ  =  99-54  grammes, 

since  the  equation  shows  us  that  73  grammes 
of  hydrochloric  acid  are  required  to  produce  44 
grammes  of  carbon  dioxide. 

If  it  be  required  to  calculate  how  much  calcium 
chloride  is  obtained  from  85  grammes  of  calcium 
carbonate,  we  have 

^5  X  lU  =  94-34  grammes, 

since  the  equation  shows  that  1 1 1  grammes  of  cal- 
cium chloride  are  obtained  from  100  grammes  of 
calcium  carbonate.  It  is  easy  to  calculate  also  how 
much  hydrochloric  acid  would  be  required  for  this 
operation,  and  how  much  carbon  dioxide  would  be 
obtained. 


14 


ARITHMETIC  OF  CHEMrSTRY 


In  the  equation 

MnOa     +     2NaCl     4-     2  H2SO4     = 

(55  +  32)  2(23  +  35-5)        2(2  +  32  +  64) 

87  117  196 

MnSO^     +     Na2S04     +     2  HgO     +     Clj 

(55  +  32  +  64)      (46  +  32  -f  64)         2(2  +  16)  71 

i5»  »42  36 

we  have  data  for  a  multitude  of  chemical  problems. 


EXAMPLES 

1.  Calculate  the  percentage  composition  of  potassium 
nitrate  (KNO3).  (Mason  Science  Coll.,  1883,  and  Columbia 
Univ.,  1895.)  This  is  equivalent  to  asking  how  much  of 
each  element  100  grammes  of  potassium  nitrate  contains, 
and  from  the  formula  it  is  known  that  loi  grammes  con- 
tain 39  grammes  of  potassium,  14  grammes  of  nitrogen, 
and  48  grammes  of  oxygen. 

Ans.  38.61%  potassium;  13.86%  nitrogen;  47.53% 
oxygen. 

2.  Calculate  the  percentage  composition  of  calcium 
sulphate  (CaSOJ.     (Columbia  Univ.,  1898.) 

Ans.  29.4%  calcium;  23.5%  sulphur;  47.1%  oxygen. 

3.  Calculate  the  percentage  of  oxygen  in :  {a)  water, 
(^)  manganese  dioxide,  and  (^)  potassium  chlorate.  (Prince- 
ton Univ.  School  of  Science,  1893.) 

Ans,  {a)  88.89;  iP)  36.87;  (0  39.18. 


1\ 


SIMPl.E  CALCULATIONS  OF  WEIGHTS 


15 


.53% 


cium 


:ygen. 


I 


;9.t8. 


4.  What  weight  of  oxygen  can  be  obtained  from  408 
grammes  of  potassium  chlorate?     (Mason  Coll.,  1882.) 

Ans.  160  grammes. 

6.  How  many  pounds  of  nitric  acid  can  be  obtained 
by  the  distillation  of  180  lbs.  of  nitre  with  sufficient  sul- 
phuric acid?     (Mason  Coll.,  1882.)  Am.  11 2. 3  lbs. 

^  6.  Oxide  of  mercury  is  heated:  (i)  alone,  (2)  with 
carbon,  (3)  with  hydrogen.  What  products  are  formed, 
and  what  weight  of  each  would  be  obtained,  if  10  grammes 
of  mercuric  oxide  were  used?     (Lond.  Matric,  1898.) 

(i)  oxygen,  0.741  gramme;  (2)  carbon  dioxide,  1.019  ; 
(3)  water,  0.834. 

7.  How  much  hydrogen  and  oxygen  is  contained  in  180 
kilogrammes  of  water.     (Princeton  Univ.,  1893.) 

Ans.  20  kilogrammes  hydrogen  ;    160  kilogrammes  oxygen. 

8.  Calculate  the  percentage  of  carbon,  hydrogen,  and 
oxygen  in  a  compound  of  the  formula  CiaH^./)!,.  (Lond. 
Matric,  1890.) 

Ans.  Carbon,  42.1%  ;  hydrogen,  6.4%  ;  oxygen,  51.5%. 

9.  100  grammes  of  hydrochloric  acid  are  mixed  with 
100  grammes  of  dissolved  caustic  soda.  What  will  be  the 
weight  of  the  compounds  contained  in  the  resulting  solu- 
tion? (The  Yorkshire  Coll.,  1890.)  There  is  more  acid  than 
is  required  to  neutralize  the  soda,  hence  there  will  be  both 
sodium  chloride  and  hydrochloric  acid. 

Ans.  Sodium  chloride,  146  grammes;  hydrochloric  acid, 
8.75  grammes. 


i6 


ARITHMET/C  OF  CHEMISTRY 


10.  If  100  parts  of  barium  chloride  give  112.1  of  barium 
sulphate,  what  is  the  atomic  weight  of  barium  ?  (The  York- 
shire Coll.  Scholarship  Exam.,  1895.) 

0=16  S  =  32  CI  =  35.45 

This  problem  is  really  to  calculate  the  value  of  x  in  the 
equation 

BaClj     -f     H2SO4     =     BaS04     +     2  HCl 
X  -I-  70.9  ^  +  32  +  64 

and  we  are  told  that  the  ratio  of  the  chloride  to  sulphate 
is  100  :  112. 1. 


Therefore 


X  -f  70.9     _   100 
^  +  32  +  64      112. 1 


,  whence  .v  =  136.7.  A?is. 


11.  If  75  c.c.  of  a  solution  of  hydrochloric  acid  are 
neutralized  by  60  c.c.  of  a  solution  of  sodium  hydroxide 
containing  0.003  gramme  of  the  alkaH  per  cubic  centi- 
metre, what  weight  of  the  acid  is  contained  in  i  c.c.  of 
its  solution?     (Cornell  Univ.,  December,  1897.) 

60  c.c.  of  alkahne  solution  contains  60x0.003  =  0.18 
gramme  of  sodium  hydroxide,  which  requires 


36.5 
40 


X  0.18  =  0.1642  gramme  of  acid. 


wJiich  is  contained  in  75  c.c. ;    therefore  i   c.c.  contains 
0.00^19  gramme  hydrochloric  acid. 

12.  A  solution  of  potassium  hydroxide  contains  0.02 
gramme  of  alkali  per  cubic  centimeter,  and  15  c.c.  of  it 
exactly  neutralize  40  c.c.  of  a  solution  of  hydrochloric  acid. 


\» 


SIMPLE  CALCULATIONS  Or  W EIGHTS        1 7 

How  many  grammes  of  hydrochloric  acid  are  contained  in 
15  c.c.  of  the  acid  solution?     (Cornell  Univ.,  1898.) 

Ans.  0.07-3  gramme. 

13.  A  solution  of  potassium  hydroxide  contains  o.oi 
gramme  of  alkali  per  cubic  centimeter,  and  ic  c.c.  of  it 
exactly  neutralize  40  c.c.  of  a  solution  of  hydrochloric  acid. 
How  many  grammes  of  hydrochloric  acid  are  contained  in 
15  c.c.  of  the  acid  solution?  Ans.  0.0244  gramme. 

14.  What  weight  of  chlorine  can  be  produced  from  100 
grammes  of  common  salt  by  the  reaction 

2  NaCl4-MnO,+  2  H^SO^^Na^SO^-t-MnSC-f  Cl,+  2  H,0? 
(Cornell  Univ.,  September,  1895.)         Ans.  60.7  grammes. 
c 


CHAPTER  III 


MORE  COMPLEX  CALCULATIONS  OF  WEIGHTS 

More  complex  problems  than  those  already  given 
may  present  themselves. 

Problem.  —  How  many  grammes  of  potassium  chlo- 
rate are  required  to  produce  enough  oxygen  for  the 
complete  combustion  of  lo  grammes  of  hydrogen  ? 

Two  chemical  reactions  are  spoken  of  in  this  prob- 
lem, and  it  is  necessary  to  know  the  nature  of  each 
reaction,  and  to  be  able  to  express  it  by  an  equation. 
The  hydrogen  is  to  be  burned,  and  enough  oxygen 
must  be  supplied  for  the  purpose.  This  oxygen  is  to 
be  obtained  from  potassium  chlorate,  of  which  a 
requisite  quantity  must  be  provided.  In  the  first 
place,  then,  we  have  to  consider  that  lo  grammes  of 
hydrogen  are  to  be  burned ;  that  is,  are  to  unite  with 
oxygen  to  form  water. 

The  equation 


2H2  + 

0,     = 

=       2  H2O 

2x2) 

32 

(2  X  18) 

4 

36 

I 

m 


18 


COMPLEX  CALCUUITIONS  OF  WEIGHTS 


19 


\ 


supplies  the  data  necessary  for  determining  the  quan- 
tity of  oxygen,  for  we  are  told  by  it  that  4  grammes 
of  hydrogen  require  32  grammes  of  oxygen ;  10 
grammes  of  hydrogen  therefore  require 

10  X  ^4^  =  80  grammes  of  oxygen. 

The  oxygen  is  obtained  from  potassium  chlorate  as 
represented  by  the  equation 


2KCIO3     =     2KCI     + 

2(39+35-5+48)       2(39  +  35.5) 
245  149 


3O2 

(3x32) 
96 


which  shows  that  to  produce  96  grammes  of  oxygen, 
245  grammes  of  potassium  chlorate  must  be  taken. 
But  the  problem  stipulates  for  the  providing  of  80 
grammes  of  oxygen,  therefore  245  x  |g  =  204.2 
grammes  of  potassium  chlorate  ^  are  required. 
Another  problem  of  the  same  nature. 

Problem.  —  How  much  carbon  dioxide  can  be  set 
free  from  marble  by  the  hydrochloric  acid  obtained 
from  30  grammes  of  common  salt.-^  The  solving  of 
this  problem  involves  the  knowledge  of  how  hydro- 
chloric acid  is  obtained  from  common  salt ;  and  also 
of  how  hydrochloric  acid  acts  on  marble.     When  sul- 


^  See  Appendix  B. 


20 


ARITHMETIC  OF  CHEMISTRY 


phuric  acid  is  put  upon  common  salt  and  the  mixture 
strongly  enough  heated,  the  reaction  takes  place  in 
the  manner  shown  by  the  equation 

2NaCl     +     H2SO4     =     2HCI     +     Na2S04 

2(23+35-5)        (2+32+64)         2(1+35.5)         (46+32+64) 
117  98  73  142 

1 1 7  grammes  of  common  salt  produce  73  grammes 
of  hydrochloric  acid,  therefore  30  grammes  of  com- 
mon salt  (the  amount  given  in  the  problem)  will 
produce 

iTY  >^  73  =  i8-7  grammes  of  hydrochloric  acid. 
The  equation 

CaCOg    +    2  HCl    =    CaClg    +    H2O    4-    CO2 

(40+12+48)      2(1+35.5)       (40+7O        (2+16)      (12+32) 
100  73  III  18  44 

shows  that  73  grammes  of  hydrochloric  acid  set 
free  44  grammes  of  carbon  dioxide;  therefore,  18.7 
grammes  of  hydrochloric  acid  would  set  free 

18  7 

— ^  X  44  =  1 1.27  grammes  of  carbon  dioxide. 

n 

These  equations,  as  put  down  above,  give  a  great 
many  more  data  than  are  necessary  for  the  problem 


COMPLEX  CALCULATIONS  OF  W EIGHTS      21 

in  hand.  It  is  advisable  for  the  beginner  to  write 
out  equations  in  full  with  the  numerical  values  for 
each  term ;  but  when  familiarity  with  the  process  is 
gained  it  is  much  neater  to  see  what  is  needed  and  to 
make  no  more  calculations  than  are  necessary.  In 
the  last  problem  the  only  substances  whose  quantity 
is  referred  to  are  carbon  dioxide,  hydrochloric  acid, 
and  common  salt ;  and  though  the  quantity  of  hydro- 
chloric acid  is  referred  to,  we  are  not  asked  to  deter- 
mine that  quantity,  though  we  did  so  in  working  out 
the  problem.  This  is  not  necessary,  however,  as  may 
be  seen  by  a  consideration  of  the  two  equations 

2  NaCl    +    H2SO4   =    2  HCl    +    Na2S04 
"7  73 

CaCOg    +    2  HCl    =    CaClg    +    HgO    -f-    CO2 

44 


These  two  equations  as  just  written  show  that  117 
grammes  of  common  salt  produce  73  grammes  of 
hydrochloric  acid,  and  that  73  grammes  of  hydro- 
chloric acid  acting  on  marble  produce  44  grammes 
of  carbon  dioxide ;  so  that  1 1 7  grammes  of  sodium 
chloride  produce  enough  hydrochloric  acid  to  liberate 
44  grammes  of  carbon  dioxide.     Thirty  grammes  of 


22 


ARITHMETIC  OF  CHEMISTRY 


common  salt  will  therefore  produce  enough  hydro- 
chloric acid  to  liberate  '^ 

44  X  i\y=  11.27  grammes  of  carbon  dioxide, 

which  is  the  same  result  as  already  obtained. 
Another  example  of  the  same  nature. 

Problem.  —  How  much  potassium  must  act  on 
water  in  order  to  produce  enough  hydrogen  to  com- 
bine with  6  grammes  of  oxygen  ? 

Potassium  acts  on  water  according  to  the  equation 


2  K     +     2  H2O 

(2  X  39) 
78 


=     2KOH     -f     H, 


and  hydrogen  combines  with   oxygen  according  to 
the  equation 


2H2     -f- 

(2x2) 

4 


O2     = 

32 


2H2O 


The  problerr  may  be  worked  out  at  full  length,  but 
by  inspection  of  the  second  equation  it  may  be  seen 
that  32  grammes  of  oxygen  require  4  grammes  of 
hydrogen ;  and  by  inspection  of  the  first  equation, 
that  in  order  to  produce  2  grammes  of  hydrogen. 


COMPLEX   CALCULATIONS  OF  WEIGHTS      23 

78  grammes  of  potassium  are  required,  and  therefore 
to  produce  4  grammes  of  hydrogen  78  x  2  =  1 56 
grammes  of  potassium  are  required. 

156  grammes  of  potassium  are  therefore  required 
to  produce  enough  hydrogen  to  combine  with  32 
grammes  of  oxygen,  and  hence,  in  order  to  produce 
enough  hydrogen  to  combine  with  6  grammes  of 
oxygen  156x^2  =  29.25  grammes  of  potassium  are 
required. 

It  is  evident  that  the  above  problem  might  be 
enlarged  by  the  inquiry  how  much  hydrogen  would 
be  left  over  if  36  grammes  of  potassium  had  been 
used,  and  it  would  merely  remain  to  calculate  how 
much  hydrogen  is  produced  by  the  action  on  water 
of  36  —  29.25  =  6.75  grammes  of  potassium. 

Problem.  —  How  much  nitrogen  is  produced  by 
the  complete  combustion  of  the  ammonia  obtained 
by  the  action  of  caustic  potash  on  15  grammes  of 
ammonium  chloride  "i 

Ammonia  burns  according  to  the  equation 


4NH3     +     3O2     =     2N2     +     6H2O 

(2  X  28) 

56 


24 


ARITHMETIC  OF  CHEMISTRY 


and  is  produced  from  ammonium  chloride  according 
to  the  equation 

NH4CI    4-     KOH    =    KCi    +    H2O    +     NHg 

(14  +  4+35.5) 

53-5 

The  first  equation  shows  that  56  grammes  of 
nitrogen  are  produced  by  the  combustion  of  four 
times  the  formula  weight  of  ammonia.  The  second 
equation  shows  that  53.5  grammes  of  ammonium 
chloride  are  required  to  yield  the  formula  weight  of 
ammonia,  and  therefore  53.5x4  =  214  grammes  of 
ammonium  chloride  are  required  to  yield  four  times 
the  formula  weight  of  ammonia.  Therefore  214 
grammes  of  ammonium  chloride  give  56  grammes 
of  nitrogen ;  1 5  grammes  of  ammonium  chloride 
will  therefore  give 

# 

56  X  gW  =  3-93  grammes  of  nitrogen. 

In  these  equations  we  have  set  down  the  formula 
weights  of  two  substances  only.  Of  these  the  prob- 
lem gives  the  actual  weight  used  of  one,  and  asks 
for  the  actual  weight  obtained  of  the  other. 

One  who  has  had  very  considerable  experience 
in  working  problems  would   see  that  the   problem 


COMPLEX  CALCULATIONS  OF  WEIGHTS     2S 


rding 


NH< 


es   of 
four 
econd 
onium 
ght  of 
nes  of 
*  times 
e   214 
mmes 
loride 


rmula 

prob- 

asks 

Hence 
>blem 


above,  expressed  in  its  simplest  terms,  involves 
merely  the  question  how  much  nitrogen  is  contained 
in  15  grammes  of  ammonium  chloride,  because  the 
second  equation  shows  that  all  of  the  nitrogen  in 
ammonium  chloride  is  obtained  as  ammonia ;  and 
the  first  equation  shows  that  all  of  the  nitrogen  in 
ammonia  is  set  free  in  the  combustion.  For  solving 
this  problem  we  need  merely  use  the  calculation 
weights  in  the  formula  of  ammonium  chloride,  and 
see  that  53.5  grammes  of  ammonium  chloride  contain 
14  grammes  of  nitrogen,  and  therefore  15  grammes 
of  ammonium  chloride  contain 

14  X  — ^=  3.93  grammes  of  nitrogen, 

the  same  result  as  already  obtained. 

The  problem  may  involve  still  more  chemical  re- 
actions and  become  still  more  complicated.  (Such 
complicated  problems,  though  seldom  set  in  exami- 
nations, are  a  splendid  exercise  for  the  student.) 

Problem.  —  Ten  grammes  of  sulphur  are  burned  in 
oxygen,  and  the  sulphur  dioxide  produced  entirely 
dissolved  in  water.  How  much  manganese  dioxide 
is  required  to  produce  enough  chlorine  to  oxidize 
the  sulphur  dioxide  to  sulphuric  acid  } 


26 


ARIlHMEriC  Oh'  CHEMISTRY 


'■  I    ; 


Here  three  distinct  chemical  reactions  are  in- 
volved, and  we  need,  therefore,  three  equations  to 
provide  the  necessary  data.  Sulphur  burns  in  oxy- 
gen according  to  the  equation 

S     4-     Oo     =     SO, 


32 


(32  +  32) 
64 


Chlorine  acts  on  sulphur  dioxide  in  water  accord- 
ing to  the  equation 

+     SO,    -f-     CI,     =     2HCI     +     H2SO4 


2HaO 


SO, 

64 


CI2     =     2  HCl 
7» 


and  manganese  dioxide  produces  chlorine  according 
to  the  equation 


MnO,    +    4  HCl 

(55  +  32) 
87 


MnCl,    +     2H2O     -H    CI2 

71 


The  equations  show  that  32  grammes  of  sulphur  pro- 
duce 64  grammes  of  sulphur  dioxide,  which  require 
for  conversion  into  sulphuric  acid,  71  grammes  of 
chlorine,  which  are  produced  by  the  action  of  hydro- 
chloric acid  on  8y  grammes  of  manganese  dioxide. 
Since  32  grammes  of  sulphur  require  87  grammes 
of  manganese  dioxide,  10  grammes  of  sulphur  will 
require 

8y  X  ^.5  =  27.19  grammes  of  manganese  dioxide. 


are  in- 
ions  to 
in  oxy- 


accord- 

H2SO4 

ccording 

+    CI2 

|hur  pro- 
require 
imes  of 
hydro- 
I  dioxide. 
Irammes 
lur  will 

)xide. 


COMPLEX   CALCULATIONS  OF   W EIGHTS     2 J 

EXAMPLES 

1.  'I'hc  hydrogen  obtained  from  6.55  graninies  of  zinc 
(acted  on  by  acid)  is  passed  over  heated  mercuric  oxide. 
What  weight  of  water  is  produced  and  how  much  mercury  ? 

Ans.  1.8  grammes  water;  20  grammes  mercury. 

2.  What  weight  of  potassium  nitrate  is  required  to  pro- 
duce enough  nitric  acid  to  make  10  grammes  of  cupric 
nitrate?  Ans.    10.77  grammes. 

3.  How  much  calcium  carbonate  could  be  made  from 
the  carbon  dioxide  obtained  by  heating  15  grammes  of 
oxalic  acid  (anhydrous)  with  strong  sulphuric  acid? 

Ans.  16.67  grammes. 

4.  What  amount  of  ammonium  nitrate  would  produce 
enough  nitrous  oxide  for  the  complete  combustion  of  36 
grammes  of  carbon?  Ans.  480  grammes, 

.5.  If  10  grammes  of  potassium  permanganate  were  dis- 
tilled with  a  concentrated  solution  of  hydrogen  chloride, 
and  the  evolved  gas  were  passed  into  a  solution  of  sulphur 
dioxide,  what  amount  of  sulphuric  acid  would  be  produced  ? 
(Lond.  Univ.  Inter.  Sci.,  1893.) 
The  equations  are 

2  KMnO^  +  16  HCl  =  2  KCl  -f  2  MnClg  +  8  H,0  +  5  CI2 
and  2H2O  +  SOo  +  CL  =  2  HCl  +  H0SO4 
These  equations  show  that  2  x  158  =  316  grammes  of  per- 
manganate provide  enough  chlorine  to  produce  5  x  98  =  490 
grammes  of  sulphuric  acid,  therefore  15.5  grammes  sul- 
phuric acid  is  the  amount  produced  by  the  10  grammes 
of  permanganate. 


m 


CHAPTKR  IV 


THE  VOLUME  OF  (JASES 


While  it  is  possible  to  make  calculations  for  the 
wcigJits  of  solids,  liquids,  and  ^ascs  involved  in  a 
chemical  reaction,  it  is  also  possible  to  make  cal- 
culations for  the  volumes  of  gases,  though  not  for 
the  volumes  of  liquids  and  solids. 

As,  however,  the  volume  of  a  gas  changes  with 
its  pressure  and  temperature,  it  is  necessary  to  state 
always  at  what  pressure  and  temperature  the  gases 
are  measured.  Before  attempting  to  make  calcula- 
tions, it  will  be  necessary  to  consider  the  manner  in 
which  pressure  and  temperature  affect  the  volume. 


Effect  of  Pressure 

Sir  Robert  Boyle,  in  the  year  1662,  showed  that 
when  the  temperature  of  a  gas  remains  constant, 
the  volume  varies  inversely  as  the  pressure.  If  a 
quantity  of  air,  for  example,  has  a  volume  of  10  gal- 
lons or  10  cubic  feet  or  10  litres,  at  the  ordinary 
atmospheric  pressure,  it  would  have  a  volume  of  5 

28 


w 


for  the 

ccl  in  a 

ike  cal- 

not  for 

jes  with 
'  to  state 
le  gases 
calcula- 
anner  in 
)lume. 


ived  that 
:onstant, 
le.      If  a 

f  lo  gal- 
ordinary 
ime  of  5 


I 


THE  VOLUME  OF  GASES 


29 


gallons  or  5  cubic  feet  or  5  litres,  at  2  atmos- 
pheres' pressure,  and  20  gallons  or  20  cubic  feet  or 
20  litres,  at  one-half  of  an  atmosphere's  pressure. 
The  statement  that  the  volume  varies  inversely  as 
the  pressure  is  called  Boyle  s  Law  and  may  be 
expressed   algebraically   in    various   forms,   such   as 


Fx-^,   or    VP 


constant, 


or,  if  V  be  the  volume  at  pressure  P,  the  volume  V-^ 
when  the  pressure  is  P^  will  be  such  that 

P  * 


F/^i  =  VP,   or    \\  =  V 


J\ 


It  is  evident  that  if  any  three  of  the  four  quanti- 
ties are  given,  the  remaining  one  may  be  calculated. 
Pressure  may  be  measured  in  atmospheres,  or  it 
may  be  measured  in  pounds  per  square  inch,  or 
it  may  be  measured  by  the  height  of  a  column 
of   mercury,   since   a   liquid   exerts  a  pressure  pro- 

*  It  is  evident  that  these  three  forms  mean  the  same  thing,  because 
V  <x.—  means  that  as  the  volume  increases  the  pressure  diminishes  in 

the  same  ratio,  and  vice  versa.  Since  the  one  factor  increases 
exactly  in  the  same  ratio  as  the  other  factor  decreases,  the  product  of 
the  two  must  always  remain  constant  ;  and  since  this  is  so  always,  it 
must  be  true  in  the  particular  case  where  the  volume  changes  from  V 
to  Vi  owing  to  the  change  of  pressure  from  P  to  Pi. 


.la 


1 1 


30 


ARITHMETIC  OF  CHEMISTRY 


portional  to  its  height.  This  last  method  's  the  most 
common  in  chemical  work.  Atmospheric  pressure 
is  ordinarily  taken  as  equal  to  the  pressure  which 
is  exerted  by  a  column  of  mercury  30  inches  or 
760  mm.  high.  A  pressure  of  36.5  inches  is  there- 
fore more  than  an  atmosphere,  and  a  pressure  of 
700  mm.  is  less  than  an  atmosphere.  We  usually 
say  the  gas  is  under  a  pressure  of  so  many  milli- 
metres. 

Problem.  —  If  a  quantity  of  oxygen  has  a  volume 
of  1 5  litres  at  a  pressure  of  760  mm.,  what  would  be 
its  volume  under  a  pressure  of  800  mm..''  (The  full 
form  of  expression  is,  "  under  a  pressure  equal  to 
that  exerted  by  a  column  of  mercury  760  mm.  high 
or  800  mm.  high,"  but  the  shorter  form  is  usually 
employed.) 


Here 


F=  15         /•=  760        /'i=  800 


Therefore     \\=  15  x  \%%  =  14.25  litres. 

In  the  calculation  one  does  not  need  to  trouble 
with  the  formula,  but  merely  ask  oneself  whether 
the  volume  will  be  greater  or  less  under  the  new 
conditions.     If  the  volume  of  the  oxygen  is  15  litres 


T}IE   VOLUME  OF  GASES 


31 


he  most 
pressure 
c  which 
ichcs  or 
s  there- 
>sure  of 
usually 
[y  milli- 

volume 
^ould  be 
rhe  full 
;qual  to 
m,  high 

usually 


trouble 
A^hether 
he  new 
[5  litres 


at  760  mm.  pressure,  it  is  evident  that  it  will  be  less 
at  a  pressure  of  800  mm.  in  the  ratio  of  ^gj{ :  i  ;  the 
volume  then  at  800  mm.  will  be  15  x  \%%, 

Problem.  —  If  the  volume  of  a  quantity  of  hydro- 
gen is  25  litres  with  a  pressure  of  720  mm.,  what 
pressure  will  the  gas  have  when  the  volume  is 
30  litres } 

Here  it  is  evident  that  the  pressure  will  be  less, 
since  the  gas  is  allowed  to  expand.  The  pressure 
will  therefore  be 

720  X  If  =  600  mm. 

Problem. — What  volume  must  carbon  monoxide 
have  had  at  a  pressure  of  650  mm.  if  its  volume 
is  18  litres  at  a  pressure  of  5cx)  mm..? 

Evidently  the  volume  must  have  been  less  at  650 
mm,  than  at  500  mm.,  so  that  we  have 

18  X  |g|=  13.84  litres. 

Effect  of  Temperature 

The  volume  of  a  gas  is  also  affected  by  the  tem- 
perature, becoming  greater  as  the  temperature  rises, 
and  less  as  the  temperature  falls. 


ii 


32 


ARITHMETIC  OF  CHEMISTRY 


ii 


The  volume  ivS  proportional  to  the  temperature,  pro- 
vided we  measure  the  tem])erature  in  a  certain  way. 

It  is  hardly  to  be  expected  that  we  could  use  the 
ordinary  thermometric  scales,  however,  because  the 
zero  in  the  centigrade  and  Reaumur  scales  is  fixed 
merely  by  the  temperature  at  which  water  Jiappcns 
to  freeze,  and  in  the  Fahrenheit  scale  the  zero  is 
still  more  accidental.  In  all  of  these  scales  we  have 
temperatures  below  zero,  which  shows  that  the  scales 
are  arbitrary.^       \ 

The  law  that  governs  the  change  of  volume  with 
temperature  could  be  determined  by  taking  a  long 
glass  tube  closed  at  one  end,  and  graduated  from  thC' 
closed  end  toward  the  open  end.  Suppose  the  tube 
graduated  into  500  equal  parts.  When  the  whole 
tube  and  the  contained  air  is  at  0°  C,  insert  a  small 
globule  of  mercury,  and  by  making  a  passage  for 
the  enclosed  air  by  means  of  a  platinum  wire  passed 
through  the  mercury,  cause  the  latter  to  move  down 
the  tube  till  it  reaches  the  273  mark.  The  mercury 
now  shuts  in  a  volume  of  air  which  occupies  273 
divisions  of  the  tube,  the  pressure  being  that  of  the 
atmosphere  (if  the  weight  of  the  mercury  is  negli- 

^  See  Appendix  C. 


THE   VOLUME  OF  GASES 


33 


gible,  or  if  the  tube  be  held  horizontally)  and  the 
temperature  being  o°  C. 

If  the  tube  be  heated  till  the  temperature  is  i°  C, 
the  mercury  will  be  driven  outward  till  it  reaches  the 
mark  274,  where  it  will  remain  so  long  as  the  tem- 
perature stays  at  1°  C.  If  the  tube  be  heated  to 
2°  C,  the  mercury  will  stand  at  the  mark  275  ;  at 
10°  C,  the  volume  of  the  enclosed  air  will  be  283,  the 
mercury  standing  at  the  mark  283.  At  the  tempera- 
ture 100°  C.  the  volume  occupied  by  the  air  will  be 
373  divisions,  the  mercury  standing  at  the  mark  373. 
When  the  temperature  reaches  227°  C,  the  mercury 
will  be  just  driven  out  from  the  tube,  the  air  having 
expanded  to  fill  the  500  divisions. 

On  the  other  hand,  if  the  tube  had  been  cooled 
down  to  —  10°  C,  the  volume  of  the  air  would  have 
been  found  to  be  263  divisions ;  if  cooled  down  to 
-  25°  C,  the  volume  would  have  been  248  divisions. 
In  general,  for  every  degree  rise  or  fall  of  tempera- 
ture, the  air  which  occupied  the  volume  of  273  divi- 
sions at  0°  C.  expands  or  contracts  through  one 
division  in  the  tube.  If  this  law  holds  to  the  very 
lowest  tem])eratures,  it  follows  that  at  —  200°  C.  the 
volimie  of  the  air  would  be  73  ;    at    —  250""  C.  the 


34 


ARITHMETIC  OF  CHEMISTRY 


W\  \ 

i 
I 


li 


volume  would  be  23  ;  and  at  —  273''  C.  the  volume 
would  be  reduced  to  zero.  This  temperature  is  called 
the  absolute  zero  of  the  air  thermometer;  and  if 
temperatures  are  measured  from  this  zero  instead 
of  from  the  artificial  zeros  of  the  ordinary  thermom- 
eters, the  volume  of  the  air  is  proportional  to  its 
temperature.  This  is  usually  called  Cliarles's  Law, 
and  is  expressed  in  the  following  form : 

If  the  pressure  remain  constant,  the  volume  of  air, 
or  any  gas,  is  proportional  to  the  temperature  meas- 
ured from  the  absolute  zero. 

The  zero  of  the  ordinary  centigrade  thermometer 
is  273°  above  the  absolute  zero,  and  therefore  when- 
ever temperatures  are  given  in  degrees  centigrade, 
273  degrees  must  be  added  in  order  to  arrive  at  the 
absolute  temperature.  If  T  represent  the  absolute 
temperature,  Charles's  Law  may  be  expressed  by  the 
formula 


Foe  7",  or 


T 


constant ; 


or  if  V  be  the  volume  at  temperature  7",  the  volume 
V-^  when  the  temperature  is  7\  will  be  such  that 


T^      T  ^  T 


1 


THE   VOLUME  OF  GASES 


35 


Problem.  —  If  the  volume  of  a  quantity  of  carbon 
monoxide  at  o°  C.  is  i8  litres,  what  will  be  its  volume 
at  25°  C. ? 

0°  C.  =  273°  absolute  temperature, 
25°  C.  =  273  +  25  =  298°  absolute  temperature ; 

therefore,  7^=273,     7^1  =  298, 

and  hence  Fj  =  18  x  |||  =  19.64  litres. 

Just  as  in  the  case  of  pressures,  instead  of  follow- 
ing the  formula  slavishly,  we  may  ask  ourselves,  will 
the  volume  be  greater  or  less  under  the  new  conditions 
than  under  the  old  }  In  this  case,  we  see  the  volume 
must  be  greater,  therefore  we  multiply  18,  the  original 
volume,  by  |||.  This  common  sense  way  of  looking 
at  the  question  will  always  prevent  mistakes. 

Problem.  —  What  must  the  temperature  be  so  that 
the  gas  which  occupies  a  volume  of  15  litres  at  20°  C. 
will  occupy  a  volume  of  16.5  litres.'*  The  tempera- 
ture, 20°  C,  is  273  -F  20  =  293°  absolute,  therefore 
7^=293.  The  new  temperature  must  evidently  be 
greater  than  this  because  the  gas  expands  from  15 
litres  to  16.5  litres.     Therefore 

7\  =  293  X  — —  =  322.3°  absolute  =  49°  3  C. 


36 


ARITHMETIC  OF  CHEMISTRY 


It  is  evident  that  the  pressure  and  temperature  of 
a  gas  may  change  simultaneously,  and  it  is  neces- 
sary to  be  able  to  calculate  the  volume  under  these 
circumstances. 

This  is  done  by  a  combination  of  Boyle  s  Law  and 
Charles's  Law. 

Boyle's  Law  gives   VP  =  const. 

Ciiarles's  Law  gives  —  =  const. 

These  two  combined  give 

VP  ,  FP 

—  =  const,  or  — 


V^ 


where  V,  P,  and  T  are  the  original  volume,  pressure, 
and  absolute  temperature,  and  Fj,  Pj,  and  T^  are 
the  final  volume,  pressure,  and  temperature.  If  any 
five  of  these  values  are  given,  the  sixth  may  be  cal- 
culated. 

[Note.  —  If  the  mathematics  given  above  is  not 
clear,  the  following  proof  may  be  read : 

Let  Vy  P,  and  T  be  the  original  volume,  press- 
ure, and  absolute  temperature  of  the  gas  Fj,  P^ 
7\,  the  final  volume,  pressure,  and  absolute  tempera- 
ture. Let  us  assume  that  the  volume  changes,  first 
owing  to  change   of   pressure  from  P  to  Pj,  the 


I 


THE  VOLUME  OF  GASES 


37 


temperature  remaining  constant  at  T.  Call  the  vol- 
ume, under  these  new  circumstances,  v\  then  by 
Boyle's  Law 


are 


I 


Let  now  the  temperature  change  from  T  to  7\ ; 
the  volume  will  change  from  v  to  V^  because  V^ 
is  the  volume  that  the  gas  occupies  when  both  press- 
ure and  temperature  are  changed. 


-.Z". 


By  Charles's  Law  Vi=  v 


T 


y  —  X  — 1 
I\      T 


or 


T 


Problem.  —  The  quantity  of  chlorine  which  occu- 
pies a  volume  of  12  litres  at  720  mm.  and  10°  C. 
has  the  pressure  changed  to  780  mm.  and  the 
temperature  to  20°  C.     What  is  the  new  volume  } 


Here 


F=  12 


P  —  720 


Pj  =  780 


r=  273 -t- 10=  283     7^1  =  273  +  20  =  293 

Therefore,  by  applying  the  formula,  we  find 
Fi  =  12  X  11^  X  III  =11.46  litres. 


38 


ARITHMETIC  OF  CHEMISTRY 


In  working  such  examples,  instead  of  following 
the  formula  slavishly,  we  may  arrive  at  the  correct 
arrangement  of  the  factors  by  a  little  consideration 
of  the  conditions.    This  may  be  seen  in  the  following 

Problem.  —  At  what  temperature  must  a  gas  be, 
so  that  its  volume  will  be  1 5  litres  when  the  press- 
ure is  800  mm.,  if  its  volume  is  17.5  litres 
when  its  temperature  is  100°  C,  and  the  pressure 
700  mm. } 

The  temperature  100°  C.  =  273  +  100=  373°  abso- 
lute ;  we  have  now  to  consider  what  the  new  tem- 
perature will  be. 

The  temperature  of  the  gas  when  its  volume  is 

15   litres  would  be  less  than  when  its  volume  was 

17.5   litres  (other  things  being   equal).      Therefore, 

if  the  volume  alone  be  considered,  the  temperature 

would  be 

15 


373  X 


17.5 


But  the  gas  in  the  final  condition  is  to  be  under 
greater  pressure  than  at  the  beginning,  and  therefore 
would  require  a  higJier  temperature  if  the  volume 
had  remained  constant,  therefore  we  must  multiply 
by  |g  J,  which  is  the  ratio  of  the  two  pressures. 


THE  VOLUME  OF  GASES 


39 


Therefore  T^  =  373  x  -^^  >^  7—  =  365°.4     ^ 
365°.4-273°  =  92°.4C. 


EXAMPLES 

1.  If  a  quantity  of  nitrogen  under  900  mm.  pressure  at 
20°  C.  occupies  a  volume  of  300  c.c,  what  volume  will 
it  occupy  at  100°  under  600  mm.  pressure?  (Cornell  Univ., 
September,  1895.)  Ans.  573  c.c 

2.  What  would  be  the  volume  under  standard  conditions 
of  a  mass  of  oxygen  whose  volume  is  60  c.c.  at  40°  C.  under 
a  pressure  of  750  mm.  of  mercury?  (Cornell  Univ.,  Decem- 
ber, 1897.)  Ans.  51.67  c.c. 

3.  If  1000  c.c.  of  chlorine  at  40°  C.  stand  in  a  tube  over 
mercury,  the  level  within  the  tube  being  30  mm.  above  that 
without,  and  the  barometric  pressure  being  750  mm.,  what 
would  be  the  volume  of  the  gas  under  standard  conditions  ? 
(Cornell  Univ.,  September,  1896.) 

Since  the  level  of  mercury  in  the  tube  is  30  mm.  above 
that  without,  the  pressure  is  750  —  30  =720  mm. 

Ans.  826.4  c.c. 

4.  If  2000  c.c.  of  nitrogen  at  27°  C.  stand  in  a  eudio- 
meter over  mercury,  the  level  within  the  tube  being  30  mm. 
below  that  without,  and  the  barometric  pressure  being 
750  mm.,  what  would  be  the  volume  of  the  gas  under 
standard  conditions?     (Cornell  Univ.,  December,  1896.) 

Pressure  in  the  tube  is  750  +  30  =  780  mm. 

Ans.  1868  c.c. 


^I'l 


m 


If: 


m 


40 


ARITHMETIC  OF  CHEMISTRY 


5.  If  1250  c.c.  of  nitrogen  at  37°  C.  stand  in  a  eudio- 
meter over  mercury,  the  level  within  the  tube  being  28  mm, 
above  that  without,  and  the  barometric  pressure  being 
747  mm.,  what  would  be  the  volume  of  the  gas  under 
standard  conditions?  Ans.  104 1  c.c. 

If  the  gas  had  been  standing  over  water  instead  of  mer- 

28  \ 


cury,  the  pressure  in  the  tube  would  have  been  (  747 


13-6, 


mm.,  since  i  mm.  of  mercury  exerts  the  same  pressure  as 
13.6  mm.  of  water. 

6.  If  a  quantity  of  hydrogen  occupies  500  c.c.  in  a  tube 
over  mercury,  the  level  within  the  tube  being  40  mm.  below 
that  without,  the  temperature  being  20°  C,  and  the  baro- 
metric pressure  730  mm.,  what  volume  will  it  occupy  at 
25°  C.  if  the  barometric  pressure  is  750  mm.  and  the  level 
within  the  tube  is  30  mm.  above  that  without?  (Cornell 
Univ.,  September,  1898.)  Ans.  544  c.c. 

7.  If  a  quantity  of  air  occupy  500  c.c.  in  a  tube  over 
mercury,  the  level  within  the  tube  being  70  mm.  above  that 
without,  the  temperature  being  40°  C.  and  the  barometric 
pressure  740  mm.,  what  volume  will  it  occupy  at  20°  C, 
the  barometric  pressure  being  720  mm.  and  the  level  within 
the  tube  25  mm.  below  that  without?  (Cornell  Univ., 
September,  1897.)  ^^•^'  4^1  c.c. 

8.  1. 041  litres  of  hydrogen,  measured  under  standard 
conditions,  is  placed  in  a  tube  over  mercury,  the  temperature 
being  37°  C.     If  the  volume  of  the  gas  is  now  1250  c.c,  at 


THE   VOLUME  OF  GASES 


41 


what  height  must  the  mercury  be  if  the  barometric  pressure 
is  755  mm.?  Ans.  36  mm. 

9.  If  186.8  c.c.  of  air  under  normal  conditions  expand 
to  2  litres  and  the  level  of  the  mercury  inside  the  tube  is 
40  mm.  below  that  without,  the  barometer  standing  at  740 
mm.,  what  must  be  the  temperature?  Ans.  27°  C. 

10.  If  a  quantity  of  moist  nitrogen  has  a  volume  25  c.c. 
at  a  pressure  of  750  mm.  and  temperature  16''  C,  what 
would  be  its  volume  when  dried,  the  temperature  remaining 
the  same,  but  the  pressure  changing  to  700  mm.?  The 
pressure  of  water  vapour  at  16°  C.  is  13.5  mm.  (see  Appen- 
dix F);    therefore  the  pressure  of  the  nitrogen  alone  is 


750  -  13.5  =  736.5  mm.,  and  25  x 


736.5 
700 


=  26.3  c.c.  Ans. 


11.  If  60  c.c.  of  dry  oxygen  at  25°  C.  has  enough  water 
put  in  to  saturate  the  gas,  what  will  be  the  volume  of  the 
moist  gas,  the  pressure  remaining  at  the  atmospheric  press- 
ure, 750  mm.?  If  the  gas  were  not  allowed  to  expand,  its 
pressure  would  become  750  +  23.5  ;  therefore,  if  allowed 
to  expand  so  that  the  pressure  will  remain  the  same,  its 


volume  must  increase  to  60  x 


773-5 
750 


=  61.9  c.c. 


12.  What  volume  would  a  quantity  of  air,  when  dried 
and  measured  over  mercury,  under  standard  conditions, 
have,  if  it  occupied  a  volume  of  100  c.c.  measured  over 
water,  and  fully  saturated  with  aqueous  vapour,  the  height 
of  the  water  in  the  tube  being  27.2  cm.,  the  temperature 
being  20°  C,  and  the  barometer  height  750  mm.? 


42 


ARITHMETIC  OF  CHEMISTRY 


27.2  cm.  of  water  equals  2  cm.  or  20  mm.  of  mercury; 
therefore  the  pressure  of  the  moist  air  is  730,  and  of  the 
dry  air  712.5,  since  the  pressure  of  the  water  vapour  is 
17.5  mm.  (nearly). 

The  required  volume  is  therefore 

712.5      273 

100  X  —7—  X  —  =  87.36  c.c. 
760       293        '  ^ 

13.  If  a  glass  tube  one  square  centimetre  in  cross- 
section  be  filled  with  mercury  and  inverted  over  a  mercury 
trough,  dry  air  passed  in  till  the  mercury  stands  at  a  divi- 
sion of  the  tube  2p  cm.  from  the  closed  end,  and  30  mm. 
above  the  mercury  outside,  the  barometer  being  750  mm. 
and  the  temperature  20°  C,  what  would  be  the  volume 
of  the  air  under  standard  conditions?  (The  mean  coefifi- 
cient  of  expansion  of  mercury  between  0°  C.  and  30°  C. 
is  0.00018143.) 

Tiiis  question  evidently  involves  making  a  correction  for 
the  expansion  of  mercury,  otherwise  the  data  would  not 
be  given.  The  barometer  reading  will  be  higher  than  it 
would  be  if  the  mercury  were  at  0°,  and  so  will  be  the 
column  of  mercury  in  the  tube.  The  pressure  on  the  gas 
will  not  therefore  be  exactly  750  —  30=720  mm.,  but  a 
less  number  for  mercury  at  0°  would  not  have  so  great 
a  height.  The  coefficient  of  expansion  of  mercury  being 
0.00018143,  means  that  the  expansion  of  each  millimetre 
is  that  amount  for  every  degree  rise  in  temperature,  and 
therefore  0.0036286  for  20°,  and  a  millimetre  length  at  0° 
would  become  1.0036286  mm.  at  20°.    A  column  of  mer- 


THE  VOLUME  OF  GASES 


43 


:ury ; 
if  the 
)ur  is 


cross- 
lercury 
a  divi- 
;o  mm. 

0  mm. 
volume 

1  coeffi- 
30°  C. 


720 

cury  720  mm.  long  at  20"*  would  therefore  be *^ --  = 

^  '  ^  1.0036286 

717.4  mm.  long  at  0°.  717.4  mm.  is  therefore  the  actual 
pressure  of  the  gas,  and  the  problem  resolves  itself  into 
a  calculation  of  the  volume  of  a  gas  at  0°  and  760  mm., 
its  volume  being  2oc.c.  at  20°  and  717.4  mm.  Ans.  17.6  c.c. 

14.  What  is  the  actual  pressure  on  the  gas  in  a  tube, 
the  level  of  mercury  inside  and  outside  the  tube  being  the 
same  when  the  barometer  reads  760  mm.  at  25°  C.  ? 

Ans.  756.6. 


\ 


ion  for 
Id  not 
;han  it 
be  the 
;he  gas 
but  a 
great 
f  being 
limetre 
re,  and 
h  at  0° 
)f  mer- 


CHAPTER   V 


CALCULATIONS  INVOLVING  WEIGHT  AND  VOLUME 


:! 


The  formula  of  a  substance  always  denotes  a  cer- 
tain definite  iveight  of  that  substance,  for  instance, 
NHg  denotes  a  weight  of  ammonia  17;  it  may  be 
pounds  or  tons  or  ounces  or  grammes  or  kilogrammes. 
In  the  same  way  QO^^^  denotes  a  weight  44,  of  carbon 
dioxide,  and  HCl  denotes  a  weight  of  36.5  of  hydro- 
chloric acid. 

But  it  is  quite  evident  that  this  definite  zveight  of 
ammonia  will  have  a  definite  fixed  volume^  if  the  gas 
be  measured  always  at  the  same  temperature  and 
pressure,  so  that  NHg  represents  a  certain  definite 
volume. 

17  pounds  of  ammonia  will  occupy  a  certain  defi- 
nite number  of  cubic  feet  when  measured  at  0°  C.  and 
760  mm.  pressure;  17  tons  of  ammonia  will  occupy 
a  larger  volume  under  the  same  conditions,  but  17 
tons  of  ammonia  will  occupy  the  same  volume  to- 
morrow as  it  does  to-day,  provided  the  pressure  and 
temperature  remain  the  same. 


44 


WEIGHT  AND   VOLUME 


45 


In  chemicai  calculations  we  do  not  ordinarily  use 
pounds  or  tons  as  the  unit  of  weight,  nor  cubic  feet 
as  the  unit  of  volume,  but  grammes  as  the  unit  of 
weight,  and  litres  as  the  unit  of  volume. 

17  grammes  of  ammonia  at  0°  C.  and  760  mm. 
occupy  the  volume  22.4  litres  (more  exactly  22.38 
litres),  and  the  formula  NHg  represents  22.4  litres 
of  ammonia  at  the  standard  temperature  and  press- 
ure, quite  as  readily  as  it  represents  17  grammes  of 
ammonia. 

In  the  same  manner  44  grammes  of  carbon  dioxide, 
represented  by  the  formula  COg,  occupy  a  certain 
definite  volume  at  the  standard  temperature  0°  C. 
and  760  mm.  This  volume  is  22.4  litres;  so  also 
36.5  grammes  of  hydrochloric  acid,  represented  by 
the  formula  HCl,  occupy  a  certain  definite  volume 
at  the  standard  temperature  and  pressure.  This 
volume  is  22.4  litres. 

In  general,  though  the  formulae  of  different  gases 
represent  different  weights,  they  always  represent 
the  same  volume  at  the  standard  temperature  and 
pressure,  and  this  volume  is  22.4  litres.  This  is 
sometimes  called  the  gramme-molecular  volume,  but 
so  far  as  calculations  are  concerned,  it  need  be  con- 


i 


■3 


46 


ARITHMETIC  OF  CHEMISTRY 


sidered  only  as  the  volume  which  is  occupied  by  the 
formula  weight  in  grammes^  quite  independently  of 
the  existence  or  non-existence  of  molecules. 

Now  tzvo  grammes  of  hydrogen  occupy  the  volume 
22.4  litres  at  the  standard  temperature  and  pressure ; 
therefore  if  the  formula  of  hydrogen  is  to  represent 
this  fact,  it  must  be  written  Hg,  for  H  represents  only 
ofte  gramme  of  hydrogen. 

In  the  same  way  thirty-two  grammes  of  oxygen 
occupy  the  volume  22.4  litres,  and  if  the  formula  of 
oxygen  is  to  represent  this  fact,  it  must  be  written 
Og,  since  O  represents  only  sixteen  grammes. 

On  the  other  hand,  200  grammes  of  mercury 
vapour,  measured  at  the  standard  temperature  and 
pressure,  would  occupy  22.4  litres ;  therefore  if  the 
formula  of  mercury  is  to  represent  this  fact,  it  must 
be  written  Hg,  since  Hg  represents  2CX)  grammes.^ 

Still  further :  1 24  grammes  of  phosphorus,  when 
volatilized,  occupy  a  volume  corresponding  to  22.4 


*  Of  course  mercury  vapour  cannot  be  made  to  exert  a  pressure  of 
760  mm.  at  o°  C,  but  if  heated  till  it  boils  it  would  have  a  pressure  of 
760  mm.  (the  pressure  of  the  atmosphere),  and  the  volume  of  200 
grammes  at  that  temperature  would  be  the  same  as  that  of  32  grammes 
of  oxygen  at  that  temperature,  and  32  grammes  of  oxygen  at  the  stand- 
ard temperature  and  pressure  has  the  volume  22.4  litres. 


WEIGHT  AND    VOLUME 


A7 


litres,  and  therefore  if  the  formula  of  phosphorus  is 
to  represent  this  fact,  it  must  be  written  P^,  since  P 
represents  only  31  grammes. 

Sulphur  a  little  above  its  boiling  point  has  a  vol- 
ume such  that  its  formula  should  be  Sg ;  at  a  very 
high  temperature  it  is  represented  by  the  formula  S^- 

The  formula  of  a  gas  must  always  represent  a 
volume  22.4  litres  at  the  standard  temp  vrature  and 
pressure. 

Ho  represents  2  grammes  of  hydrogen  and  occupies  a  volume  22.4  litres 


02 

a 

32 

a 

oxygen 

a 

<( 

22.4   " 

N2 

n 

28 

ii 

nitrogen 

^i 

^^ 

22.4    " 

C12 

t( 

71 

n 

chlorine 

n 

(t 

22.4   " 

03 

«( 

48 

it 

ozone 

ti 

ti 

22.4    " 

P4 

a 

124 

ik 

phosphorus 

(( 

n 

22.4    " 

AS4 

<-■ 

300 

>< 

arsenic 

«< 

u 

22.4    " 

Hg 

^| 

200 

u 

mercury 

(( 

(( 

22.4    " 

HCl 

<t 

36.5 

a 

hydrochloric  acid 

« 

n 

22.4    " 

CO 

(( 

28 

i. 

carbon  monoxide 

tt 

•• 

22.4    " 

COo 

« 

44 

.. 

carbon  dioxide 

(• 

i( 

22.4    " 

CHt 

(« 

16 

(( 

methane 

(( 

X 

22.4    " 

C2H2 

n 

26 

a 

acetylene 

<« 

it 

22.4    " 

CcHe 

(( 

78 

(( 

benzol 

i( 

(( 

22.4    " 

CoHcC 

»" 

46 

(( 

alcohol 

(( 

<( 

22.4    " 

The  relative  weight  of  the  different  gases  can 
easily  be  seen  from  the  above  table.  For  instance, 
chlorine  is  JJ  times  as  heavy  as  oxygen,  wjiile  carbon 
monoxide  has  the  same  density  as  nitrogen.      Air, 


48 


AR/THMKT/C  OF  CHEMISTRY 


which  is  made  up  of  oxygen  and  nitrogen,  has  a  den- 
sity lying  between  them,  being  nearer  that  of  nitro- 
gen, since  this  element  forms  the  larger  portion  of  the 
atmosphere.  22.4  litres  of  air  weigh  28.8  grammes. 
The  specific  gravity  of  any  of  the  gases  (that  is,  its 
density  compared  with  air)  is  therefore  obtained  by 
dividing  its  formula  weight  by  28.8.  Thus  arsenic 
vapour  is  more  than  10  times  as  heavy  as  air;  that 
is,  its  specific  gravity  is  greater  than  10. 

Acetylene  and  benzol  have  the  same  percentage 
composition  which  could  be  represented  by  the  for- 
mula CH  ;  but  if  this  were  the  formula  of  either  of 
them,  13  grammes  would  occupy  the  volume  22.4 
litres.  Such  is  not  the  case,  however :  26  grammes 
of  acetylene  occupy  the  volume  22.4  litres,  therefore 
its  formula  is  C2H2 ;  and  78  grammes  of  benzol  vapour 
occupy  the  volume  22.4  litres,  and  therefore  CgHg  is 
the  formula  for  benzol.  There  is  abundant  addi- 
tional evidence  in  favour  of  these  formulae,  but  for 
the  purpose  of  calculation  nothing  more  is  required. 

Facts  similar  to  the  ones  given  above  caused 
Avogadro  to  make  the  statement  in  181 1  that  equal 
volumes  of  gases  at  the  same  temperature  and  press- 
ure contain  equal  numbers  of    molecules,  and   this 


fess- 
this 


WEIGHT  AND    VOLUME 


49 


statement  is  usually  called  Avogadros  Laiv.  It  was 
neglected  for  a  long  time,  because  chemists  did  not 
have  a  clear  conception  of  the  facts  at  the  foundation 
of  it,  and  the  facts  known  were  not  so  numerous  or 
varied  as  at  present. 

So  far  as  concerns  calculations  Avogadro's  Law 
might  be  expressed :  The  formula  weight  in 
grammes,  of  a  gas  measured  at  o°  C.  and  760  mm., 
occupies  a  volume  of  22.4  litres. 

Problem.  —  What  volume  of  carbon  dioxide  is  got 
from  the  combustion  of  10  litres  of  carbon  monoxide, 
and  what  volume  of  oxygen  is  required } 

The  necessary  equation  is 

2  volumes         i  volume  2  volumes 

2  X  22.4  litres     22.4  litres     2  X  22.4  litres 

2  CO     +     Oa     =     2CO2 

(It  is  convenient  to  write  weights  beneath  the  for- 
mulae and  volumes  above.) 

The  equation  shows  that  the  volume  of  the  carbon 
dioxide  is  exactly  the  same  as  that  of  the  monoxide, 
and  that  the  volume  of  the  oxygen  is  half  as  great ; 
and,  therefore,  in  the  case  given,  lo  litres  of  carbon 
dioxide  would  be  produced,  and  5  litres  of  oxygen 
would  be  required  for  the  combustion. 


50 


ARITHMETIC  OF  CHEMISTRY 


Problem.  —  What  volume  of  hydrochloric  acid  is 
obtained  when  6  litres  of  hydrogen  unite  with 
chlorine  ? 

The  equation  is 


I  volume       I  volume 
22.4  litres       22.4  litres 

H,     +     CL     = 


2  volumes 
2  X  22.4  litres 

2HCI 


The  volume  of  the  hydrochloric  acid  is  therefore 
double  that  of  the  hydrogen ;  therefore,  in  the  case 
given,  12  litres  of  hydrochloric  acid  would  be 
obtained. 

Since  the  formula  represents  iveight  as  well  as 
volume^  the  weight  of  one  substance  might  be  given, 
and  the  volume  of  another  required,  or  vice  versa. 
(It  is  of  course  to  be  remembered  that  the  volume 
of  gases  only  can  enter  into  the  problem.) 

Problem.  —  What  volume  of  oxygen  at  the  normal 
temperature  and  pressure  can  be  obtained  from  10 
grammes  of  potassium  chlorate.?     The  equation  is 


2  KCIO3     = 

2(39  +  35-5  +  48) 
245 


2KCI 


3  volumes 
3  X  22.4  litres 

+     3  0j 


WEIGHT  AND    VOLUME 


51 


showing  that  245  grammes  of  chlorate  yield 
3  X  22.4  =  67.2  litres  of  oxygen.  Therefore,  10 
grammes  of  chlorate  will  yield 


67.2  X  ^^^  =  2.74  litres. 

We  now  see  why  we  should  not  write  the  equa- 
tion in  the  form  KCIO3  =  KCl  4-  O3.  This  would 
mean  that  when  122.5  grammes  of  potassium  chlo- 
rate are  heated,  22.4  litres  of  ozone  are  obtained, 
which  is  not  true. 

Nor  should  it  be  written  KCIO3  =  KCl  +  3  O, 
because  O2  and  not  O  represents  the  formula  weight 
of  oxygen  corresponding  to  the  standard  volume. 

Problem.  —  What  weight  of  ammonium  nitrate 
will  provide  10  litres  of  nitrous  oxide  at  the  normal 
temperature  and  pressure  (0°  C.  and  760  mm.).? 

From  the  equation 


NH4NO3     =     2  H2O 

(14  -I-  4  +  14  +  48) 
80 


+ 


I  volume 
22.4 

N2O 


it    is    seen    that   22.4   litres    of    nitrous    oxide    are 
obtained   from   80  grammes  of   ammonium  nitrate; 


52 


ARITHMETIC  OF  CHEMISTRY 


therefore,    lo  litres  of    nitrous  oxide  are  obtained 
from 

80  X =  35.71  grammes  of  ammonium  nitrate. 

22.4 

.  Since   ammonium    nitrate    is   a   solid,   its  volume 
cannot  be  calculated. 


Problem.  —  What  volume  of  hydrogen  measured 
at  0°  C.  and  700  mm.  can  be  obtained  by  the  action 
of  sulphuric  acid  on  10  grammes  of  zinc  ? 

The  equation 

I  volume 
22.4  litres 

Zn     +     H2SO4     =     Zn  SO4  4-     Hg 
65.5 

shows  that  65.5  grammes  of  zinc  yield  22.4  litres 
of  hydrogen  at  the  standard  temperature  and  press- 
ure, 0°  C.  and  760  mm. ;  therefore  10  grammes  of 
zinc  will  under  these  conditions  yield 

22.4  X  -^  =  3.42  litres. 
65.5 

In  the  problem  the  temperature  is  the  standard 
temperature,  but  the  pressure  is  not  the  standard 
pressure,  so  that  a  correction  must  be  made  for  the 
pressure.      Since    the    pressure,    700  mm.,    in    the 


WEIGHT  AND    VOLUME 


53 


tained 

rate, 
olume 


asured 
action 


le 
es 


|.  litres 

press- 

nes  of 


mdard 
mdard 
or  the 
n    the 


example  is  less  than  the  normal  pressure,  the  volume 
must  be  greater  than  that  calculated  above.  The 
volume  will,  therefore,  be 

3.42  X  ^%  =  3.71  litres. 

Problem.  —  What  weight  of  copper  is  required  to 
produce  15  litres  of  nitric  oxide  at  800  rnm.  and 
20°  C. } 

Since  equations  give  the  relation  between  weight 
and  volume  at  the  standard  pressure  and  tempera- 
ture, it  is  necessary  first  of  all  to  calculate  the 
volume  of  the  nitric  oxide  at  this  tiormal  pressure 
and  temperature. 

The  normal  (or  standard)  pressure,  760  mm.,  is 
less  than  that  at  which  the  gas  is  measured,  namely, 
800  mm. ;  and  therefore  the  volume  at  760  mm. 
will  be  f  g^  times  as  great. 

The  standard  temperature  is  lower,  and  on  that 
account  the  volume  would  be  less.  The  normal 
(or  standard)  temperature  is  273°  absolute  tem- 
perature, while  that  at  which  the  gas  is  measured 
is  273  -I-  20  =  293°  absolute,  and  therefore  the  vol- 
ume at  the  normal  temperature  (temperature  alone 
being   considered)  is   Hf    times   as   great.      When 


54 


ARITHMETIC  OF  CHEMISTRY 


:  I 


change  of  pressure  and  change  of  temperature  are 
both  considered,  we  have  therefore 

volume  =  15  X  fj^  x  ||f  =  14.71  litres. 

To  arrive  at  the  weight  of  copper  necessary  to 
produce  this  volume  of  nitric   oxide  the  equation 

2  volumes 
2  X  22.4 

3  Cu  +  8  HNO3  =  3  Cu(N08)2  +  4  H2O  4-  2  NO 

3  X  63.5 
190.5 

is  used,  and  shows  that  to  provide  2  x  22.4  =  44.8 

litres  of  nitric  oxide  190.5  grammes  of  copper  are 

required;  therefore  to  produce  14.71  litres  of  nitric 

oxide 

190.5  X -^^  =  62.52   grammes  of   copper  are 

required. 


44.8 


Problem.  —  What  volume  of  nitrogen,  measured 
at  20°  C.  and  750  mm.,  would  be  obtained  by  the 
action  of  ammonia  on  300  c.c.  (cubic  centimetres) 
of  chlorine  measured  at  15°  C.  and  735  mm..? 

This  corresponds  to  a  possible  change  of  tem- 
perature and  pressure  in  a  laboratory  from  one  time 
of  measuring  a  gas  to  another.  A  cubic  centimetre 
is  y-Q^QTr  ^^  ^  \i^^Qy  but  as  we  are  dealing  with  volumes 
merely,  the  problem  is  to  be  worked  out  in  the  same 


WEIGHT  AND    VOLUME 


55 


are 


are 


tem- 

Itime 

letre 

imes 

ame 


way  as  though  we  were  dealing  with  litres,  but  the 
answer  ol^tained  is  in  cubic  centimetres,  not  in  litres. 
The  equation  to  represent  the  reaction  is 


8NH. 


3  X  22.4 
+      3CI2 


22.4 
6NH,C1     +     N, 


and  shows  that  three  times  22.4  litres  of  chlorine 
yield  22.4  litres  of  nitrogen  at  the  normal  tempera- 
ture and  pressure ;  or  three  times  22.4  c.c.  of  chlorine 
will  yield  22.4  c.c.  of  nitrogen ;  or,  in  general,  the 
volume  of  the  chlorine  is  tJiree  times  that  of  the 
nitrogen  produced  by  it. 

Therefore  300  c.c.  of  chlorine  will  yield  ^^ 
=  100  c.c.  of  nitrogen  if  the  two  are  measured  at 
the  same  temperature  and  pressure,  since  all  gases 
are  equally  affected  by  changes  in  temperature  and 
pressure,  and  the  same  relationship  that  holds  at 
the  normal  temperature  and  pressure  will  hold  at 
any  other.  We  have  then  to  determine  what  vol- 
ume will  be  occupied  by  a  quantity  of  nitrogen 
at  20°  C.  and  750  mm.  if  its  volume  is  100  c.c.  at 
I5°C.  and  735  mm. 
F=ioo    r=273+ 15  =  288°    7\  =  273  +  20  =  293° 

P=735  ^1=750 

therefore     Fj  =  100  x  |||  x  {|f  =  99.77  c.c. 


56 


ARnHiMETIC  OF  CHEMISTRY 


The  increase  in  pressure  is  nearly  counterbalanced 
by  the  rise  in  temperature. 

Problem.  —  What  weight  of  ammonia  would  be 
required  in  the  last  problem  ? 

Here  we  must  remember  that  equations,  as  we 
have  learned  to  look  upon  them,  show  the  rela- 
tionship between  grammes  and  litres  at  o°  C.  and 
760  mm. 

22.4  litres 


N 


2 


8NH3     +     3CI2     =     6NH4CI     -f- 

8  (14  +  3)  grammes 
136 

shows  that  136  grammes  of  ammonia  are  required  to 
produce  22.4  litres  of  nitrogen  at  0°  C.  and  760  mm. 
But  since  a  cubic  centimetre  is  loW  °^  ^  \\tx^,  iVo\ 
gramme  or  0.136  gramme  or  136  milligrammes  of 
ammonia  would  be  required  to  produce  22.4  c.c. 
of  nitrogen  measured  at  0°  C.  and  760  mm.  But 
the  problem  asks  how  much  ammonia  is  required 
to  produce  100  c.c.  of  nitrogen  measured  at 
15°  C.  and  735  mm.;  therefore  we  must  find  out 
what  this  volume  will  become  at  0°  C.  and  760  mm. 
It  will  evidently  be  less  because  the  temperature 
is  less ;  and  less  also  because  the  pressure  is  greater. 


WEIGHT  AND    VOLUME 


57 


be 


We  have  therefore 


volume  =  icx)  x  |J|  x  {§J  =  91.68  c.c. 

Our  problem  is  ultimately  reduced  to  the  question : 
What  weight  of  ammonia  is  required  to  produce 
91.68  c.c.  of  nitrogen  when  136  mg.  of  ammonia 
produce  22.4  c.c.  of  nitrogen } 

The  answer  is  evidently 

136  X  ^^^—  =  556.5  mg.  =  0.5565  gramme. 
22.4 

Problem.  —  At   what   temperature   is   the   oxygen 

measured  if  10  litres  of  it,  at  a  pressure  of  740  mm., 

is   sufficient   for    the    complete    combustion    of    10 

grammes  of  phosphorus } 

The  equation 

5  X  22.4 
5O2     =     2P2O6 


*4 
124 


■f 


shows   that    124    grammes    of    phosphorus    require 
112  litres  of  oxygen  at  0°  C.   and  760  mm. 
Therefore  10  grammes  of  phosphorus  require 

1 12  X  -^^  =  9.04  litres. 

The  question  resolves  itself  into :    At  what  tem- 
perature  will   9.04   litres    of    oxygen   (measured   at 


■fVk 


I 


58 


ARITHMETIC  OF  CHEMISTRY 


the  normal  temperature  and  pressure)  expand  to 
10  litres,  its  pressure  being  reduced  to  740  mm.  ? 

Since  the  volume  is  to  increase,  the  temperature 
should  on  that  account  be  higher,  but  since  the 
pressure '\^  to  be  lessened,  so  high  a  temperature 
would  not  be  required  as  would  otherwise  be  the 
case. 

The  fraction  giving  the  volume  factor  is  there- 
fore greater  than  unity,  and  the  fraction  giving  the 
pressure  factor  is  less  than  unity ;  therefore 

T^  =  273  X  -'^-  X  249  =  294.5  =  2i°.5  C. 
9.04     700 

It  is  evident  that  questions  involving  the  use  of 
two  or  three  equations  may  arise  in  calculations  of 
volumes  as  well  as  in  calculations  of  weights,  but 
they  introduce  no  new  principles  and  need  not  be 
further  discussed. 

EXAMPLES 

1.  J50  c.c.  of  oxygen  at  10°  C.  and  756  mm.  Find  the 
volume  at  o°C.  and  760  mm.    (Mason  Science  Coll.,  1887.) 

Ahs.  240  c.c. 

2.  Find  the  amount  in  grammes  of  2500  c.c.  of  sulphur 
dioxide  measured  at  20°  and  760  mm.  (Mason  Coll., 
1887.)  Alls.  6.65  grammes. 


. 


id  the 
1887.) 
40  c.c. 

lUlphur 
Coll., 
mmes. 


WEIGHT  AND    VOLUME 


59 


3.  Calculate  the  weight  of  100  litres  of  hydrogen  sul- 
phide measured  at  27°  C.  and  740  mm.  (Lond.  Matric, 
June,  1892.)  Ans.  134.6  grammes. 

4.  What  volume  of  sulphuretted  hydrogen,  measured 
at  0°  C.  and  760  mm.  pressure,  can  be  obtained  from 
10  grammes  of  ferrous  sulphide?  (Edin.  Univ.,  ist  Pro- 
fessional Med.,  1871.)  Ans.  2.55  litres. 

5.  What  volume,  in  litres,  of  oxygen,  measured  at  740 
mm.  pressure  and  27°  C.  temperature,  can  be  obtained 
from  54  grammes  of  mercuric  oxide?  (Edin.,  ist  Profes- 
sional, 1871.)  Ans.  3.16  Htres. 

6.  State  the  relations  of  volume  in  the  gases  involved 
in  the  following  equations : 

C  +  4N2+     02=4N2  +  C02 

C  -f  4  N2  +  CO.  =  4  N2  -f  2  CO  , 

(Columbia  Univ.,  1895,  ^^^  y^'"'^'',  2d  term.) 

7.  A  piece  of  pure  marble,  weighing  20  grammes,  is 
strongly  ignited  in  a  large  porcelain  crucible.  Required 
the  name  and  weight  of  the  compound  which  remains 
behind,  and  the  name  and  volume  in  cubic  centimetres 
at  20°  C.  and  740  mm.  mercurial  pressure  of  the  gas  which 
escapes.     (Princeton  Univ.,  1896.) 

Lime  =  11.2  grammes  ;  carbon  dioxide  =  4938  c.c. 

8.  What  volume  of  oxygen,  when  burned  with  hydrogen, 
will  yield  40  litres  of  water  vapour  under  the  initial  condi- 
tions of  temperature  and  pressure  ?  Ans.  20  litres. 

Solve  by  inspection.    (Cornell  Univ.,  September,  1897.) 


6o 


ARITHMETIC  OF  CHEMISTRY 


9.    What  volume  of  olefiant  gas  would  be  equal  in  weight 
to  ID  litres  of  marsh  gas?     (Lond.  Matric,  January,  1889.) 

Ans.  5^  litres. 

10.  How  many  cubic  centimetres  of  hydrogen  can  9.2 
grammes  of  sodium  liberate  from  water,  the  barometer 
standing  at  710  mm.  and  the  thermometer  at  12°  C? 
(Edin.  Univ.,  ist  Professional,  1889.) 

Ans.  5.00  litres  =  5000  c.c. 

11.  How  many  litres  of  nitrous  oxide,  measured  at  700 
mm.  pressure  and  15°  C.  temperature,  can  be  obtained  by 
the  decomposition:  of  100  grammes  of  ammonium  nitrate? 
(Edin.  Univ.,  1st  Professional,  1871.)      Ans.  32.07  litres. 

12.  How  many  cubic  centimetres  of  hydrogen,  measured 
at  20°  C.  and  750  mm.,  will  be  obtained  by  dissolving 
3.25  grammes  of  pure  zinc  in  an  acid?  (Mason  Coll., 
1882.)  Ans.  1. 216  litres  =  1216  c.c. 

13.  Calculate  the  volume  of  vapour  measured  at  100°  C. 
that  would  be  given  by  5  c.c.  of  bromine,  the  specific 
gravity  of  the  liquid  being  2.96.     (Mason  Coll.,  1886.) 

Weight  of  bromine  =  5  X  2.96  =  14.8  grammes  ; 
therefore,  volume  required    =    2.83  litres. 

14.  What  weight  of  ammonium  nitrate  must  be  decom- 
posed under  a  barometric  pressure  of  740  mm.  at  20°  C.  to 
generate  sufficient  nitrous  oxide  to  fill  a  balloon  holding  200 
litres?     (Cornell  Univ.,  June,  1896.)       Ans.  648  grammes. 


15.    What  volume  of  carbon  dioxide  will  result  from  the 


WEIGHT  AND   VOLUME 


6l 


/eight 
889.)     ' 
litres. 

m  9.2 
)meter 
2°  C? 

00  c.c. 

at  700 
led  by 
litrate? 

1  litres. 

pasured 

solving 

Coll., 

16  c.c. 

00°  C. 
specific 
886.) 


decom- 
3°  C.  to 
ing  200 
ammes. 

om  the 


combustion  of  20  litres  of  carbon  monoxide  ?  (Solve  with- 
out using  combining  weights.)  (Cornell  Univ.,  March, 
1898.)  Ans.  20  Htres. 

16.  Explain  why  28.88  multiplied  by  the  density  of  a  gas 
gives  its  molecular  weight.     (Cornell  Univ.,  1898.) 

Molecular  weight  is  what  we  have  called  formula  weigh  . 

17.  («)  Find  the  molecular  weight  of  the  gas  whose 

density  is  1.524. 

(b)  Find  the  density  of  the  gas  whose  molecular 
formula  is  C2H4. 

{c)  A  litre  of  nitrous  oxide  weighs  1.97  grammes; 
find  its  molecular  weight.  (Cornell  Univ., 
1898.) 

{a)  Density  is  compared  to  air,  therefore  the  for- 
mula weight  =  28.8  X  1.524  =  43.9. 
28 


(0 


=  0.973. 


28.8 
1.97  X  22.4  =  44.1. 

18.  What  volume  of  chlorine  at  normal  temperature  and 
pressure  would  be  required  (a)  to  combine  with  10  litres 
of  olefiant  gas;  {b)  to  decompose  10  litres  of  sulphuretted 
hydrogen  liberating  the  sulphur ;  (c)  to  decompose  10 
grammes  of  potassium  iodide?  (London  Matric,  June, 
1889.)      {a)  10  litres;    (b)   10  litres;    {c)  0.675  litre. 

19.  How  many  litres  of  oxygen,  measured  at  15°  C.  and 
750  mm.  pressure,  are  required  for  the  complete  combustion 
of  100  grammes  of  marsh  gas?  (Edin.  Univ.,  ist  Profes- 
sional, 1890.)  Ans.  299.3  litres. 


62 


ARITHMETIC  OF  CHEMISTRY 


20.  What  volume  in  litres,  of  oxygen  measured  at  730 
mm.  and  10°  C,  can  be  obtained  from  i  kilogramme  of 
pyrolusite,  containing  80%  of  peroxide  of  manganese? 
(Edin.  Univ.,  ist  Pro^'essional,  1873.)        Ans,  74.1  litres. 

21.  What  volume  of  sulphurous  acid  gas,  measured  at 
740  mm.  pressure  and  15°  C.  temperature,  can  be  obtained 
by  burning  10  grammes  of  sulphur?     (Edin.  Univ.,  1872.) 

Ans.  7.57  litres. 

22.  {a)  Find  the  density  of  the  gas  whose  molecular 

formula  is  C2N2. 
{b)  Find  the  weight  of  a  litre  of  hydrochloric  acid. 
{c)    I   litre  of  a  gas  weighs   1.25  grammes   under 

standard  conditions ;  find  its  molecular  weight. 

(Cornell  Univ.,  September,  1896.) 
Ans.  {a)  1.807;  (^)  1.629  grammes;  (^)  27.9. 

23.  How  many  litres  of  dry  atmospheric  air  at  740  mm. 
pressure  and  15°  C.  temperature  are  required  to  burn  com- 
pletely I  litre  of  olefiant  gas  at  the  same  temperature  and 
pressure,  and  what  is  the  weight  of  carbonic  anhydride  and 
of  water  produced?     (Edin.  Univ.,  1870.) 

Take  air  as  containing  21  %  of  oxygen.  Ans.  14^  litres  air. 
Weight  of  carbonic  anhydride 

%Z  X X  ~-  X  -||  =  3-63  grammes. 

22.4      760      288     '^    '^  ^ 

Weight  of  water  =  1.486  grammes. 

24.  What  volume  of  air  at  15°  C.  and  750  mm.  would 
be  required  to  burn  to  litres  of  marsh  gas  measured  at  the 
normal  temperature  and  pressure?  (Assume  air  contains 
20.5%  of  oxygen.)    (Mason  Coll.,  1885.)    Ans.  104.1  litres. 


WEIGHT  AND   VOLUME 


63 


25.  50  c.c.  of  marsh  gas  are  mixed  with  600  c.c.  of  air 
and  exploded.  What  will  be  the  composition  and  volume 
of  the  residual  gas,  the  measurements  being  made  through- 
out at  normal  temperature  and  pressure?  (The  Yorkshire 
Coll.,  1896.)  Ans.  476  c.c.  nitrogen;  24  c.c.  oxygen; 

50  c.c.  carbon  dioxide. 

26.  What  volume  will  9  grammes  of  water  occupy  meas- 
ured in  the  form  of  vapour  at  273°  C.  and  under  380  mm. 
pressure?     (Aberdeen  Univ.,  Sci.  and  Med.,  1897.) 

Ans.  44.8  Htres. 

27.  What  volume  of  gaseous  ammonia,  measured  at  15°  C. 
and  under  380  mm.  pressure,  may  be  obtained  from  20 
grammes  of  ammonium  chloride?  (Aberdeen  Univ.,  Sci. 
and  Med.,  1898.)  Ans.  17.66  litres. 

28.  Calculate  the  weight  of  10  litres  of  air,  measured 
at  0°  C.  and  760  mm.  pressure,  containing  21%  of  oxygen 
and  79%  of  nitrogen.     (Lond.  Matric,  June,  1888.) 

tVt  X  32  +  1%^  2^  —  28.84  grammes  in  22.4  litres. 

X  28.84  =  12.88  grammes.  Ans. 

22.4 

29.  If  hydrogen  were  passed  over  100  grammes  of 
heated  copper  oxide  as  long  as  action  took  place,  what 
would  be  the  quantities  of  the  resulting  substances,  and 
what  volume  of  the  gas  measured  at  the  standard  tem- 
perature and  pressure  would  be  needed  to  complete  the 
change?     (London  Matric,  June,  1893.) 

Ans.  Copper  =  79.9  grammes;  water  =  22.65  grammes; 
hydrogen  =  28.2  litres. 


64 


ARITHMETIC  OF  CHEMISTRY 


30.  A  cubic  metre  of  steam  measured  at  iio°C.,  and 
under  standard  pressure,  is  condensed  to  water  at  4°  C. ; 
what  would  be  the  bulk  of  the  water,  and  what  volume  of 
hydrogen  (at  0°  C.  and  760  mm.)  would  afford  such  a 
quantity  of  water  when  burned?  (London  Matric,  Jan- 
uary, 1894.)     I  c.c.  of  water  at  4°  C.  =  i  gramme. 

Ans.  573  c.c.  water;  713  litres  hydrogen. 

31.  17  grammes  of  a  mixture  of  potassium  chlorate 
and  manganese  dioxide  gave  3  litres  of  oxygen  at  20°  C. 
and  760  mm. ;  calculate  the  percentage  of  potassium  chlo- 
rate in  the  mixture.     (The  Yorkshire  Coll.,  1896.) 

Ans.  60%. 

32.  What  volume  of  air  at  15°  C.  and  740  mm.  would 
be  necessary  for  the  combustion  of  10  grammes  of  cyano- 
gen gas  ?  (The  Yorkshire  Coll.,  1895.)  Assume  air  to 
contain  20.5%  oxygen.  Ans,  45.5  litres. 

33.  A  spark  is  passed  through  a  mixture  of  750  c.c. 
of  hydrogen  and  250  c.c.  of  oxygen,  both  at  the  tempera- 
ture of  100°  C.  What  will  be  the  resulting  volume  measured 
at  100°  C?     (The  Yorkshire  Coll.,  1890.) 

Ans.  750  c.c.  consisting  of  500  c.c.  water  vapour  and 
250  c.c.  hydrogen. 

34.  What  would  be  the  weight  of  hydrogen  obtained 
from  7  grammes  of  oil  of  vitriol  containing  98.2%  of  sul- 
phuric acid?  What  volume  would  it  occupy  at  19°  C.  and 
744  mm.?     (Queen's  Coll.,  Galway,  1897.) 

Ans.  0.1402  gramme;  1.71  litres. 


WEIGHT  AND   VOLUME 


65 


35.  0.2  gramme  of  iron  wire  by  solution  in  acid  gave 
83.7  c.c.  of  hydrogen  at  i2°C.  and  750  mm.  What  per- 
centage of  pure  iron  is  contained  in  the  sample?  (Mason 
Coll.,  1882.)  Ans.  98.9%. 

36.  A  blue  flame  is  often  seen  at  the  top  of  a  coke  or 
charcoal  fire.  To  what  substance  is  it  due  and  how  is  the 
flame  formed?  Suppose  that  the  coke  used  contained  90% 
of  carbon  and  it  was  entirely  converted  into  this  substance, 
how  much  coke  would  be  needed  to  make  10,000  litres 
measured  at  15° C.  and  750  mm.?    (London  Matric,  1896.) 

Ans.  5570  grammes. 

37.  50  grammes  of  crystallized  oxalic  acid  (H2C2O4, 
2  H2O)  are  decomposed  by  heating  with  strong  sulphuric 
acid.  What  is  the  composition  of  the  gas  evolved,  and 
what  volume  (in  c.c.)  will  it  occupy  if  measured  at  15°  C. 
and  748  mm.?     (Mason  Coll.,  1882.) 

Ans.  Equal  volume  of  carbon  monoxide  and  carbon 
dioxide.     Total  volume  =  91,030  c.c. 

38.  What  volume  of  gas  measured  at  standard  tempera- 
ture and  pressure  would  be  evolved  by  the  action  of  con- 
centrated sulphuric  acid :  {a)  on  10  grammes  of  oxalic, 
{U)  on  10  grammes  of  formic  acid,  both  anhydrous?  (Aber- 
deen Univ.,  1 89 1.)  Anhydrous  oxalic  acid  has  the  formula 
C2H2O4,  and  formic  acid  the  formula  H2CO2,  and  the  latter 
decomposes  according  to  the  equation 

H2CO2  =  H2O  4-  CO. 

Ans.  (a)  4.98  litres ;  (d)  4.86  litres. 


66 


ARITHMETIC  OF  CHEMISTRY 


39.  What  volume  of  nitric  oxide  gas  measured  at  o°C. 
and  760  mm.  can  be  obtained  by  acting  on  1000  grammes 
of  ferrous  sulphate  (FeSO^,  7  HjO)  with  the  necessary  quan- 
tities of  nitric  and  sulphuric  acids?  (Edin.  Univ.,  1889.) 
The  equation  is 

6(FeS04,  7  H2O)  -V  2  HNO3  +  3  H2SO4 

=  3  Fe2(S04)3+  46  H2O  4-  2  NO    Ans.  26.86  litres'. 

40.  How  many  litres  of  nitric  oxide  gas  measured  at 
750  mm.  and  10°  C.  can  be  obtained  by  treating  100 
grammes  of  ferrous  chloride  (FeCl2)  with  excess  of  nitric 
acid?     (Edin.  Univ.,  187 1.)  ^«j.  6.17  litres. 

41.  A  solution  of  potassium  permanganate  containing 
31.6  grammes  per  litre  was  added  to  20  c.c.  of  a  solution 
of  hydrogen  peroxide  acidulated  with  sulphuric  acid  till  the 
colour  was  no  longer  destroyed.  35.8  c.c.  of  the  perman- 
ganate solution  were  used.  What  amount  of  peroxide  was 
present  per  100  c.c.  of  the  solution?  It  was  observed  that 
gas  was  given  off.  What  volume  would  100  c.c.  of  peroxide 
solution  have  afforded?   (London  Univ.  B.  Sc,  1893.)    The 

equation  is 

2  KMn04  -f  5  H2O2  +  3  H2SO4 

=  K2SO4  +  2  MnS04  +  8  H2O  +  5  O2. 

The  weight  of  permanganate  in  35.8  c.c. 

_35.8 


1000 


X  31.6=  1. 133  grammes. 


Weight  of  peroxide  in  20  c.c. 

=  1. 1 33  X  Jj^J  =  0.608  gramme. 


' 


WEIGHT  AND   VOLUME  67 

Weight  of  peroxide  in  100  c.c. 

=  0.608  X  5  =  3.04  grammes. 

Volume  of  oxygen  =  ^^  x  112  =  2.005  litres. 

42.  0.02  gramme  of  a  dyad  metal  when  dissolved  in 
hydrochloric  acid  evolved  83.81  c.c  of  hydrogen  measured 
in  the  moist  state  under  a  pressure  of  750  mm.  and  at  15°  C. 
Calculate  the  atomic  weight  of  the  metal,  vapour  press- 
ure of  water  at  15°  C.  being  12.7  mm.  (London  Univ., 
Intermediate  Exam,  in  Science,  1895.)  The  formula  of 
the  chloride  is  RClj,  where  R  stands  for  metal,  since  the 
latter  is  dyad.  The  pressure  of  hydrogen  is  750  —  12.7  = 
737.3  mm.  The  volume  of  hydrogen  under  standard 
conditions 

=  83.81x^X^  =  77.39. 
200       760 

The  problem  reduces  itself  to :  If  0.2  gramme  of  the 
metal  yield  77.39  c.c.  of  hydrogen,  how  much  will  yield 
22,400  c.c?  Ans.  58.1,  which  is  the  atomic  weight. 

43.  What  volume  of  air  would  be  required  for  the  com- 
plete combustion  of  100  litres  of  a  gas  containing  hydrogen 
46%,  marsh  gas  40%,  olefiant  gas  14%  by  volume?  (Lon- 
don Int.  Sci.,  1889.) 

Ans.  Of  air  containing  20%  oxygen,  725  litres. 

44.  One  gramme  of  a  metal  A  when  dissolved  in  acid 
liberates  930  c.c.  of  hydrogen  measured  at  18°  C.  and 
750  mm. ;    also  i  gramme  of  a  metal  B  when  dissolved 


r 


e>% 


ARITHMETIC  OF  CHEMISTRY 


in  nitric  acid  yields  1.65  grammes  of  nitrate.  From  these 
data  calculate  the  equivalent  of  each  metal.  (London 
Univ.  Int.  Sci.,  1890.)  Ans.  A  26  -,  B  103.8. 

45.  What  volume  of  air  containing  21%  of  oxygen 
measured  at  20°  C.  and  760  mm.  will  be  required  to  burn 
completely  10  grammes  of  pure  carbon?  (London  Univ., 
Prelim.  Sci.  Med.,  January,  1890.)  Ans.  95.4  litres. 

46.  What  weight  of  potassium  bichromate  would  be  re- 
quired to  provide  enough  chlorine  to  combine  with  20  litres 
of  ethylene  measured  at  20°  C.  and  770  mm.? 

'  Ans.  82.6  grammes. 

47.  How  much  calcium  carbide  would  be  required  to 
produce  enough  acetylene  to  yield,  on  combustion,  14  litres 
of  carbon  dioxide  measured  at  15°  C.  and  740  mm.,  and 
what  volume  of  oxygen  at  the  same  temperature  and  press- 
ure would  be  needed  ? 

Ans.  18.5  grammes  calcium  carbide;  17.5  litres  of  oxygen. 

48.  What  volume  of  oxygen  measured  at  10°  C.  and 
780  mm.  pressure  would  be  required  for  the  complete  com- 
bustion of  the  marsh  gas  obtainable  from  20.5  grammes 
of  sodium  acetate?  Ans.  11.3  litres. 


r 


CHAPTER  VI 


CALCULATIONS  IN   VOLUMETRIC  ANALYSIS 


It  is  not  within  the  scope  of  this  little  book  to 
treat  at  all  fully  the  subject  of  volumetric  analysis, 
but  a  few  examples  may  be  given  to  illustrate  the 
principles  set  forth  in  the  preceding  chapters. 

In  volumetric  analysis,  solutions  are  used  whose 
strength  is  known.  In  some  cases  the  strength  may 
follow  no  regular  system.  These  may  be  called  arbi- 
trary solutions,  or  unsystematic  solutions.  For  ex- 
ample, a  potassium  permanganate  solution  might  be 
made  up,  somewhat  at  haphazard,  and  be  found  by 
testing  to  have  such  a  strength  that  i  c.c.  of  it  will 
oxidize  0.01324  gramme  of  iron  from  the  ferrous  to 
the  ferric  condition ;  in  other  words,  change  the 
amount  of  ferrous  sulphate,  containing  that  much 
iron,  into  ferric  sulphate  (sufficient  sulphuric  acid 
being  assumed  present). 

It  will  readily  be  seen  that  if  a  gramme  of  ore  is 
taken,  each  c.c.  of  the  permanganate  used  will  repre- 

69 


70 


ARITHMETIC  Oh    CHEM/STRV 


sent  1.324%  of  iron  (for  it  is  0.01324  gramme  in  one 
gramme). 

To  simplify  calculations,  solutions  which  may  be 
called  cvcn-numbcr  solutions  are  sometimes  used.  For 
example,  if  a  permanganate  solution  is  prepared  so  that 
each  cubic  centimetre  equals  0.0 100  gramme  of  iron, 
then,  if  i  gramme  of  ore  require  31.2  c.c.  of  this  solu- 
tion, the  amount  of  iron  is  seen  at  once  to  be  31.2%. 
Even-number  solutions  are  the  most  convenient  ones, 
so  far  as  saving  of  calculation  goes,  in  cases  where 
only  a  single  substance  is  to  be  determined  with  the 
volumetric  solution ;  but  where  several  substances  are 
to  be  determined  with  the  solution,  the  latter  will  be 
an  even-number  solution  for  not  more  than  one  of 
the  substances.  For  instance,  a  solution  of  perman- 
ganate which  would  give  percentages  in  i  gramme 
of  iron  ore,  would  not  give  percentages  of  manga- 
nese in  I  gramme  of  a  manganese  ore  or  per- 
centages of  oxalic  acid  in  an  impure  oxalate. 

When  the  solution  is  used  for  varied  purposes,  cal- 
culations are  somewhat  simplined  b}'  the  use  of  nor- 
mal solutions,  or  solutions  which  bear  some  smiple 
numerical  relation  to  the  latter,  such  as  \  normal,  -^ 
normal  (decinormal),  or  -^-^  normal  (centinormal). 


! 


VOLUMETRIC  ANALYSIS 


71 


This  savinj^  in  calculation  is  usually  accomplished 
by  taking  a  weight  of  the  substance  for  analysis 
which  bears  a  simple  relation  to  the  standard,  or  for- 
mula, weight  of  the  substance  sought.  For  example, 
if  0.56  gramme  of  iron  ore  is  taken,  each  cubic  cen- 
timetre of  a  decinormal  potassium  bichromate  solution 
will  correspond  to  1%  of  iron,  or,  if  0.80  gramme 
of  ore  is  taken,  will  correspond  to  i  %  of  ferric  oxide 
(Fe^Oa). 

The  statement  just  made  involves  a  distinct  defi- 
nition of  the  term  "normal,"  and  since  the  term  is 
used  in  different  senses  by  different  chemists,  it  is 
always  necessary,  in  working  problems  involving  the 
term,  to  make  sure  that  the  definition  required  by  the 
problem  is  clearly  borne  in  mind. 

According  to  one  definition,  the  normal  solution 
contains  the  standard,  or  molecular,  weight  in 
grammes,  of  the  substance  in  i  litre ;  according 
to  another  definition,  the  normal  solution  contains 
the  gramme  equivalent  of  the  substance  per  litre; 
while,  according  to  a  third  definition,  the  normal 
solution  contains  the  hydrogen  equivalent  in  grammes 
per  litre.  According  to  any  of  the  definitions  the 
standard,  or   normal,  solution  of   hydrochloric  acid 


\\'\' 


72 


ARITHMETIC  OF  CHEMISTRY 


would  have  a  weight  of  36.5  grammes  of  pure  hydro- 
chloric acid  in  the  volume  of  i  litre.  According  to 
the  first  definition,  normal  sulphuric  acid  solution 
would  have  98  grammes  of  sulphuric  acid  (H2SO4) 
in  the  Utre.  But  98  grammes  of  pure  sulphuric  acid 
would  neutralize  twice  as  much  caustic  potash  as 
36.5  grammes  of  pure  hydrochloric  acid,  so  that,  if 
the  acid  solutions  are  to  be  equivalent,  the  sulphuric 
acid  should  contain  only  49  grammes  of  sulphuric 
acid,  which  is  the  normal  solution  according  to  the 
second  definition. 

The  potassium  permaixganate  solution,  which  has 
the  formula  weight  KMn04  of  salt  in  a  litre,  is  capa- 
ble of  oxidizing  five  times  the  amount  of  hydrogen 
contained  in  the  normal  solution  of  hydrochloric 
acid,  and,  according  to  the  third  definition,  the  nor- 
mal solution  would  contain  1|^  grammes  {\  KMn04) 
per  litre ;  that  is,  has  one-fifth  the  concentration  of 
the  normal  solution,  according  to  the  first  definition. 

The  second  and  third  definitions  amount  to  the 
same  thing  in  most  cases ;  but  since  there  may  be 
ambiguity,  in  some  instances,  as  to  the  signification 
of  ** equivalent  weight,"  the  term  "normal  solution" 
will,  in  this  chapter,  be  used  in  the  sense  of  the  third 
definition. 


VOLUMETRIC  ANALYSIS 


73 


ydro- 
ng  to 
lution 

2SO4) 
c  acid 
Lsh  as 
hat,  if 
phuric 
phuric 
to  the 

ch  has 
s  capa- 
drogen 
chloric 
\e  nor- 
Mn04) 
tion  of 
inition. 
to  the 
nay  be 
cation 
ution  " 
e  third 


Problem.  —  If  i  c.c.  of  hydrochloric  acid  neutral- 
izes 0.0106  gramme  of  sodium  carbonate,  what 
weight  of  silver  chloride  will  be  precipitated  when 
I  c.c.  is  added  to  excess  of  silver  nitrate  solution  ? 
(Yale  Scientific  School,  1898.) 

The  equations  are 

NaaCOg    +    2  HCl    =    2  NaCl    +    Hfi    +    COj 
AgNOg    +    HCl   =    AgCl    +    HNO3 

which  show  that  two  formula  weights  of  silver 
chloride  correspond  to  one  formula  weight  of  sodium 
carbonate. 

2  X  143.5  grammes  of  silver  chloride  are  pre- 
cipitated by  the  acid  which  is  neutralized  by  106 
grammes  of  sodium  carbonate,  or  0.0287  gramme  of 
silver  chloride  corresponds  to  0.0106  gramme  of 
sodium  carbonate,  and  would  therefore  be  pre- 
cipitated by  I  c.c.  of  the  given  hydrochloric  acid 
solution. 

Problem.  —  If  i  c.c.  acid  =1.25  c.c.  alkali  solution, 
and  if  to  standardize  50  c.c.  acid  solution,  10  c.c. 
alkali  solution  and  0.25  gramme  of  sodium  car- 
bonate are  used ;  i  c.c.  alkali  is  equivalent  to  what 
weight  of  actual  hydrochloric  acid  (HCl).'' 


m: 


li'i 


I  >> 


!   I 


74 


ARITHMETIC  OF  CHEMISTRY 


I  c.c.  acid  =  what  weight  of  nitrogen  (by  Kjeldahl)  ? 
(Yale  Scientific  School,  1898.) 

10  c.c.  alkali  solution  =  8  c.c.  acid,  so  that  42  c.c. 
acid   solution   are    neutralized    by   0.25    gramme   of 

sodium  carbonate,  or  i  c.c.  by  -^— ^  gramme. 

42 

By  the  equation  above,  73  grammes  of  hydro- 
chloric acid  neutralize   106  grammes  of  sodium  car- 

bonate,  and   hence   to   neutralize    --^    gramme    of 

42 
0.25 


sodium  carbonate,  73  x 


=  o.(X)4099   gramme 


42  X  106 
of  actual  hydrochloric  acid  is  necessary. 

By  Kjeldahl's  method  nitrogen  is  converted  into 
ammonia,  and  each  standard  weight  of  nitrogen, 
namely  14,  corresponds  to  the  standard  weight  of 
hydrochloric  acid,  36.5. 

Therefore,  the  nitrogen  is 

14  _ 


0.004099  X 


36.5 


0.001572  gramme. 


\  1 1 


i   \ 


Problem. — What  volume  of  centinormal  acid 
would  be  required  to  neutrahze  0.005  gramme  of 
potassium  carbonate  (KgCOg).'' 

The  equation 


K2CO3     +     2  HCl     =     2  KCl    +     H2O    +    CO2 


VOLUMETRIC  ANALYSIS 


75 


ahl)? 

\2  C.C. 

ne  of 

hydro- 
m  car- 

me    of 

rramme 

ted  into 
trogen, 
ight  of 


shows  that  138  grammes  of  potassium  carbonate 
are  neutralized  by  twice  the  amount  of  hydrochloric 
acid  contained  in  a  litre  of  normal  strength ;  and 
as,  according  to  the  definition  of  normal  which  we 
adopt,  all  acids  of  normal  strength  are  equivalent, 
138  grammes  of  potassium  carbonate  would  be 
neutralized  by  2  litres  of  any  acid  of  normal 
strength,  or  by  200  litres  of  acid  of  centinormal 
strength. 

To  neutralize  o.(X)5  gramme  of  carbonate, 

200  X  — — -  =  0.007242  litre  =  7.24  c.c.^ 
138  ^  ^ 

Problem.  —  How  many  cubic  centimetres  of  a  deci- 
normal  permanganate  solution  would  be  required  to 
just  produce  a  colour  in  100  c.c.  of  a  centinormal 


m 


al     acid 
mme  of 


+    CO5 


1  While  the  above  is  the  exact  interpretation  of  the  equation,  it 
is  more  common  to  express  the  fact  in  a  slightly  different  form,  and 
instead  of  saying  that  the  formula  weight  of  potassium  carbonate 
requires  tivice  the  formula  weight  of  hydrochloric  acid,  to  say  that 
half  the  formula  weight  of  potassium  carbonate  requires  the  formula 
weight  of  hydrochloric  acid,  or  that  since  the  formula  weight  of 
potassium  carbonate  contains  twice  the  standard  (or  atomic)  weight 
of  potassium,  half  of  its  formula  (or  molecular)  weight  must  be 
taken  to  be  equivalent  to  the  formula  (or  molecular)  weight  of 
hydrochloric  acid  which  contains  the  standard  (or  atomic)  weight 
of  hydrogen. 


h\U 


■       ,1    V-ff.,     .    ,j^.^J..J..^.^^^...    ■:.  ■■■ 


I  !!£ 


76 


ARITHMETIC  OF  CHEMISTRY 


I  I 


solution  of  peroxide  of  hydrogen  to  which  sulphuric 
acid  is  added  ? 
The  equation 

2KMn04     +     5H2O2     +     3H2SO4 

2MnS04     +     5O2     4-     8H2O 


=     K2SO4     + 


represents  the  fact  that  the  oxygen  comes  equally 
from  the  permanganate  and  from  the  peroxide,  and 
therefore  the  normal  solutions  are  equivalent ;  and, 
therefore,  10  c.c.  of  a  decinormal  permanganate  is 
equal  to  100  c.c.  of  a  centinormal  peroxide  solution. 
Problem. — What  must  be  the  strength  of  a 
ferrous  sulphate  solution  such  that  100  c.c.  of  it 
will  be  sufficient  just  to  decolourize  20  c.c.  of  a  per- 
manganate solution  containing  15.8  grammes  of  salt 

per  litre } 

ioFeS04  +  8H2SO4  +   2KMn04 
10  X  152  2  X  158 

=    5Fe2(S04)3  +  K2SO4  +  2MnS04  +  8  H2O 

From  this  equation  it  is  evident  that  2  x  158 
grammes  of  permanganate  oxidize  10  x  152  grammes 
of  ferrous  sulphate ;  and,  therefore,  if  the  perman- 
ganate could  oxidize  an  equal  volume  of  the  ferrous 
sulphate   solution,  the   latter  must  contain    5  x  152 


ill 


phuric 


8H2O 

equally 
de,  and 
it;  and, 
mate  is 
solution. 
1    of     a 
c.   of  it 
a  per- 
s  of  salt 


8H2O 

2  X  158 
crrammes 

perman- 
e  ferrous 
5  X  152 


VOLUMETRIC  ANALYSIS 


77 


grammes  of  ferrous  sulphate  for  every  158  grammes 
of  potassium  permanganate  in  the  permanganate 
solution.  In  the  problem  we  have  assumed  the 
ferrous  sulphate  solution  to  be  only  one-fifth  as 
strong,  and  therefore  it  must  contain  152  grammes 
of  sulphate  for  every  158  grammes  of  permanganate, 
or  15.2  grammes  per  htre.  If  we  were  dealing  with 
crystallized  ferrous  sulphate  (FeS04,  7  H2O),  the 
amount  required  would  be  27.8.  It  is  important  to 
notice  this,  since  the  salt  is  usually  weighed  out  in 
the  crystalline  form. 

Problem.  —  What  weight  of  iron  is  contained  in 
I  c.c.  of  the  ferrous  sulphate  solution  in  the  last 
question,  and  what  per  cent  of  iron  is  there  in  an 
ore  if  13.6  c.c.  of  the  permanganate  solution  be  used 
for  1 .42  grammes  of  ore  .'' 

The  iron  is  -^^.^  of  the  ferrous  sulphate ;  there- 
fore, since  there  is  15.2  grammes  ferrous  sulphate 
per  litre,  there  would  be  5.6  grammes  of  iron  per 
litre,  or  0.0056  gramme  per  i  c.c. 

The  permanganate  solution  in  th'.  given  case  is 
capable  of  oxidizing  Jive  times  its  volume  of  a 
ferrous  sulphate  solution  containing  this  quantity  of 


a. 


W 


e.ii'Uii".'y-^H'.^fj;M   W    >.HW<MirtBitfcl»i>»ifc">'ii«WI>i*»iai— K 


I 


'  ',1 
!    ! 


ill 

'si 


ii 

111 


!l! 


\i 
nil 

!  I 


m 


■:! 


78 


ARITHMETIC  OF  CHEMISTRY 


i!'! 


iron;  therefore,  13.6  c.c.  of  permanganate  solution 
would  oxidize  13.6  x  5  =68  c.c.  of  ferrous  sulphate, 
which  contains  68  x  0.0056  =  0.3808  gramme  of  iron. 
1.42  grammes  of  ore  contains  0.3808  gramme  of 
iron,  or  26.82  %. 

Problem. — What  weight  of  ore  must  be  taken  so 
that  the  number  of  cubic  centimetres  of  the  perman- 
ganate solution  above  will  represent  the  percentage 
in  iron?  j 

I  c.c.  of  the  permanganate  is  equivalent  to 
0.0056  X  5  =  0.028  gramme  of  iron,  and  if  this  is 
equal  to  I  %,  the  total  ore  must  be  2.8  grammes. 

Problem. — What  must  be  the  strength  of  a  per- 
mangate  solution  if  the  burette  reading  in  cubic 
centimetres  gives  the  percentage  of  iron  direct  when 
0.7  gramme  of  ore  is  taken .? 

In  this  case  i  c.c.  of  permanganate  must  corre- 
spond to  0.007  gramme  of  iron,  since  if  the  substance 
were  pure  iron,  100  c.c.  of  the  permanganate  would 
be  necessary  for  the  0.7  gramme;  and  since  15.8 
grammes  of  the  permanganate  correspond  to 
5.6  X  5  =  28  grammes  of  iron,  the  permanganate 
solution    required    by    the    problem    must    contain 


VOLUMETRIC  ANALYSIS 


79 


per- 
cubic 
when 

corre- 
stance 
would 

|e   15-8 

d     to 

anate 

ontain 


I  >%  8 
0-7  X  -^^=  0.395  gramme  permanganate  per  100  c.c, 

or  3.95  grammes  per  litre. 

Problem.  —  i  c.c.  of  a  permanganate  solution  is 
equivalent  to  (will  oxidize)  0.014  gramme  of   iron; 

1  c.c.  is  equivalent  to  what  weight  of  manganese  in 
a  hot  neutral  solution  (Volhard's  method).?  (Yale 
Scientific  School,   1898.) 

This  reaction  goes  according  to  the  equation 

3MnS04     4-     2KMn04     +     2  HgO 
=     sMnOg     +     K2SO4     +     2H2SO4 

The  equation  which  represents  the  reaction  of  per- 
mangariat<;  with  a  ferrous  salt  shows  that  2  KMn04 
oxidizes    10  Fe;     while    this    equation    shows    that 

2  KMn04  oxidizes  3  Mn.  Therefore  the  amount 
of  permanganate  which  would  oxidize  10x56=560 
grammes  of  iron  would  oxidize  3  x  55  =  165  grammes 
of  manganese. 

The  answer  to  the  problem  is  therefore  |||  x 0.014 
=  0.0041  gramme  of  manganese. 

Problem.  —  If  10  c.c.  of  a  permanganate  solution 
are  heated  with  hydrochloric  acid,  the  liquid  evap- 


m 


Vi. 


ialiiUMii 


80 


ARITHMETIC  OF  CHEMISTRY 


\  ill 


i 


;  -.1, 


ii! 


'ill 

■1: 
ill 

ij! 

■  :i 

11! 


%     % 


orated  with  excess  of  sulphuric  acid,  the  residue 
taken  up  with  water,  neutralized  with  zinc  oxide, 
and  boiled,  how  many  cubic  centimetres  of  the 
same  permanganate  solution  will  be  required  to  pre- 
cipitate  the   manganese?      (Yale   Scientific    School, 

1899) 
The   question   gives   the   details   of    the   Volhard 

process,  and  the  equation  in  the  last  problem  shows 
that  there  is  \  as  much  manganese  in  the  perman- 
ganate as  in  the  manganese  sulphate,  and  therefore, 
if  a  given  quantity  of  permanganate  is  changed  to 
manganese  sulphate,  it  will  require  |  as  much  per- 
manganate to  oxidize  it  again  to  the  condition  of 
dioxide ;   hence  in  this  case  6|  c.c. 

Problem.  —  A  solution  of  permanganate  contain- 
ing 6.32  grammes  per  litre  : 

{a)  What  is  its  strength  in  terms  of  FcgOs? 

{b)  What  amount  of  iron  ore  must  be  weighed 
out  so  that  each  cubic  centimetre  used  will  repre- 
sent one  per  cent  of  iron  (Fe)? 

{c)  What  is  the  strength  of  this  solution  against 
manganese  by  Volhard's  method  ?     (Columbia  Univ.) 


VOLUMETRIC  ANALYSIS 


8l 


sidue 
)xide, 
■  the 
3  pre- 
chool, 

>lhard 
shows 
jrman- 
refore, 
ged  to 
h  per- 
ion  of 


ontain- 


reighed 
repre- 

against 
,  Univ.) 


Ans.    (a)    0.0160    gramme   of    FejOg   per    cubic 
centimetre. 

(d)    1. 1 2  grammes. 

(c)    0.0033   gramme   manganese   per  cubic 
centimetre. 

Problem.  —  If  0.5  gramme  of  iron  is  dissolved, 
without  oxidation,  in  water  containing  5  grammes 
of  sulphuric  acid  (H2SO4),  and  the  requisite  amount 
of  permanganate  is  run  in,  what  weight  of  free 
acid  will  the  solution  contain  ?  (Yale  Scientific 
School,   1899.) 

By  examination  of  the  equations  it  will  be  seen 
that  in  order  to  dissolve  the  iron  and  to  complete 
the  reaction  with  permanganate,  1 8  H2SO4  is  re- 
quired   for    loFe;    therefore,    for   0.5    gramme   of 

18  X  08 
iron  ~  X  0.5    gramme    of    sulphuric   acid    are 

necessary ;  therefore  the  free  acid  is 

5  _  M  ^  98  X  0.5J  =  3.425  grammes. 

Problem. — Two  grammes  of  ore  containing  ar- 
senic were  treated  according  to  Pearce's  method, 
and  the  silver  in  the  silver  arsenate  determined 
by    titration    with    ammonium   sulphocyanate ;     the 


82 


ARITHMETIC  OF  CIIEMISTKY 


11^ 


amount  was  27.9  c.c,  the  strcnjjfth  of  solution 
I  c.c.  =  0.007  gramme  of  silver.  What  was  the 
percentage  of  arsenic } 

Pearce's  method  consists  in  obtaining  neutral  so- 
dium arsenate,  precipitating  the  arsenic  acid  as 
silver  arsenate,  collecting  the  precipitate  and  dis- 
solving in  nitric  acid,  and  titrating  with  ammonium 
sulphocyanate  solution. 

In  the  problem  given,  the  silver  used  was  27.9 
X  0.0007  =  0.1953  gramme,  and  from  the  formula 
of  the  precipitate  Ag3As04,  3  x  108  =  324  parts 
of  silver  correspond  to  75  parts  of  arsenic,  there- 
fore the  arsenic  =0.1953x^^^  =  0.0452  gramme 
in  2  grammes  of  ore;   that  is,  2.26%. 

Problem.  —  How  much  crystallized  sodium  thio- 
sulphate  (Na2S203,  5  H2O)  must  be  dissolved  in 
water  and  diluted  to  i  litre  to  give  a  solution  whose 
strength  is  i  c.c.  =0.015  gramme  of  copper.? 


2Cu(N03)2  +   4KI   =   CU2I2   +   I2  +  4KNO3 

I2  +   2Na2S203  =   Na2S40(.   +   2  Nal 

that  is,  one  standard  weight  of  copper  63.5   corre- 
sponds   to    one    formula    weight    of    thiosulphate 


/  01. UMKTRH '  ANAL  \ 'SIS 


83 


(NivS/)^,    5ir/T)    248;     therefore,   0.015  Xp-'^—  = 

0.05859   gramme   of   thiosuli)hate    per    cubic    centi- 
metre, or  58.59  grammes  per  litre. 


Problem.  —  What  weight  of  crystallized  sodium 
thiosulphate  will  make  1000  c.c.  of  a  solution, 
I  c.c.  of  which  will  be  ecpiivalent  to  0.0008  gramme 
of  oxygen  (available),  by  lUnisen's  method  of  dis- 
tilling a  higher  oxide  of  manganese  with  hydro- 
chloric acid  and  collecting  the  chlorine  in  potassium 
iodide  solution.'*     (Yale  Scientific  School,   1898.) 

0.0008  gramme  of  oxygen  corresponds  to  0.00355 
gramme  of  chlorine,  and  this  to  0.0127  gramme  of 
iodine,  and  by  the  equation  above  0.0127  gramme 
iodine  corresponds  to  0.0248  gramme  of  thiosulphate 
crystals;  therefore,  1000  c.c.  should  contain  24.8 
grammes  of  sodium  thiosulphate  crystals. 

Problem.  —  We  have  a  rich  copper  matte  and 
decide  to  take  0.5  gramme  for  analysis.  How  much 
sodium  thiosulphate  (NagS.^Os)  per  litre  will  give  a 
solution  such  that  i  c.c.  will  equal  2  %  of  copper  ? 
(Columbia  Univ.) 

From  the  equations  above,  63.5  grammes  of  cop- 


( 


f     !1 


IMAGE  EVALUATION 
TEST  TARGET  (MT-3) 


:/ 


z 


% 


1.0  ^KfiUi 

1.1  l.-^KS 

1.25  ||.4   ||.6 

J 

c" 

o       — — 

%^ 


71 


/. 


Wd 


c»^ 


Hiotographic 

Sciences 

Corporation 


<F  - 


\ 


«■ 


ff 


^  \  ^\ 


^j% 


'<«^ 


23  WEST  MAIN  STREET 

WEBSTER,  N.Y.  14S80 

(716)872-4503 


'% 


b'  *<- 


^,v 


^ 


84 


ARITHMETIC  OF  CHEMISTRY 


per  require  158  grammes  of  thiosulphate ;  hence  0.5 

gramme  of  copper  would  require  — ^  x  0.5=  1.2442 

63-5 

grammes  of  thiosulphate,  which  must  be  contained 
in  50  c.c,  since  if  the  ore  were  pure  copper,  50  c.c. 
of  the  thiosulphate  solution  must  be  added,  accord- 
ing to  the  conditions  of  the  problem. 

There  must  therefore  be  1.2442  x  20  =  24.884 
grammes  per  litre. 

Problem.  —  i  c.c.  of  a  potassium  bichromate  solu- 
tion contains  x  grammes  of  the  pure  salt,  i  c.c. 
will  liberate  what  weight  of  iodine  from  an  acid 
solution  of  potassium  iodide }  i  c.c.  will  be  equiva- 
lent to  what  weight  of  iron } 

The  equation 

KjCraO^     -f-     6KI     -h     14HCI 

=     8KC1     -h     2CrCl3     4-     3 12     +     7H2O 

shows  that  294  grammes  of  bichromate  yield  762 
grammes  of  iodine;  hence  \%\x  is  the  weight  of 
iodine. 

A  consideration  of  the  equation  will  show  that 
there  are  16  x  3  =  48  grammes  of  oxygen  useful 
for  oxidizing  purposes  in  294  grammes  of  bichro- 


VOLUMETRIC  ANALYSIS 


85 


mate ;  or,  as  is  usually  said,  there  is  3  O  (available) 
which  will  oxidize  6Fe  from  ferrous  to  ferric  con- 
dition;  hence  the  weight  of  iron  is  ^x. 

294 


Problem.  —  Suppose  i  c.c.  of  permanganate  solu- 
tion is  equivalent  to  0.028  gramme  of  iron.  What 
will  be  the  strength  of  the  oxalic  acid  solution,  such 
that  it  will  just  decolourize  an  equal  volume  of  the 
permanganate  ?  t 

Since  158  grammes  of  permanganate  oxidizes 
56  X  5  =  280  grammes  of  iron,  the  solution  of  perman- 
ganate in  the  example  must  contain  15.8  grammes 
per  litre. 

But  from  the  equation 

2  KMnO^  +  5  (C00H)2 -h  3  H2SO4 

=  K2SO4  -I-  2  MnS04  +  10  CO2  +  8  H2O 

it  is  evident  2  x  158  grammes  of  permanganate 
oxidize  5  x  90  grammes  of  oxalic  acid.  Therefore, 
the  solution  of  oxalic  acid  to  be  equivalent  to  the 
permanganate  must  contain  |x9=22.5  grammes 
oxalic  acid  per  litre. 

Problem.  —  Compare    the    phosphorus    standard 
with  the  iron  standard  for  potassium  permanganate. 


r 


86 


.11 


ARITHMETIC  OF  CHEMISTRY 


When  phosphorus  is  precipitated  as  yellow  ammo- 
nium phospho-molybdate,  the  ratio  is  one  standard 
weight  of  phosphorus  (P)  to  12  standard  weights 
of  molybdic  acid  (MoOg),  and  the  weight  of  the 
phosphorus  is  if  J^  that  of  the  molybdic  acid. 

The  molybdic  acid  is  then  reduced  -o  the  com- 
pound represented  by  the  formula  MojOg,  which  is 
oxidized  by  permanganate  according  to  the  equation 

5  MoaOa  +  6  KMnO^  +  9  HaS04 

=  10  Mopg  4-  3  K2SO4  +  6  MnS04  +  9  H3O 

so  that  6  KMn04  corresponds  to  10  MoOg. 

But  we  saw  that  6  KMn04  corresponds  to  30  Fe, 
so  that  MoOg  corresponds  to  3  Fe,  or  144  grammes 
of  molybdic  acid  correspond  to  168  grammes  of  iron. 

But  31  grammes  of  phosphorus  correspond  to 
1728  grammes  of  molybdic  acid;  therefore  to  1728 
X  {If  grammes  of  iron. 

Therefore, 

phosphorus  standard :  iron  standard 
=  31  :  1728  X  III  =  0.01538:  I, 

which  means  that,  in  the  titration  for  phosphorus, 
I  c.c.  of  permanganate  solution  represents  a  much 
smaller  amount  of  phosphorus  than  it  would  of  iron, 
in  an  iron  estimation. 


/^ 


VOLUMETRIC  ANALYSIS 


87 


lorus, 

I  much 

iron, 


Problem.  —  If  a  sample  of  iron  contains  97%  pure 
iron,  and  if  it  requires  as  much  of  the  given  perman- 
ganate sokition  in  the  phosphorus  estimation  as  in 
the  iron  estimation,  what  is  the  percentage  of  phos- 
phorus ?  31  X  97  X  144  ^      , 

Problem.  —  What  weight  of  substance  must  be 
weighed  out  for  analysis : 

(a)  so  that  i  c.c.  of  the  decinormal  acid  will  be 
equivalent  to  i  %  of  nitrogen  by  Kjeldahl  ? 

(if)  so  that  I  c.c.  of  the  same  acid  will  represent 
I  %  of  sodium  carbonate  (NagCOg)  ? 

(c)    I  %  of  sodium  oxide  (NagO).    (Yale  Scientific 

School,  1898.) 

(a)  0.14  gramme;    (d)  0.53  gramme; 

(c)  0.31  gramme. 

Problem.  —  A  quantity  of  caustic  potash  solution 
requires  32  c.c.  of  normal  acid  solution  to  neutralize 
it.  What  is  the  weight  of  potassium  hydrate  (KOH) 
in  the  solution  }  Ans.  1.792  grammes. 

Problem.  —  5.3  grammes  of  moist  sodium  carbonate 
require  90  c.c.  of  a  normal  solution  of  acid  to  neu- 
tralize it.    What  is  the  percentage  of  water } 

Ans.  10%. 


I: 


n 


r 


^^ 


ARITHMETIC  OF  CHEMISTRY 


.11 


)  Problem.  —  If  a  solution  of  silver  nitrate  contains 
17  mg.  of  salt  per  cubic  centimetre,  how  many  cubic 
centimetres  of  a  centinormal  potassium  chloride  solu- 
tion would  be  needed  to  precipitate  the  silver  con- 
tained in  9.64  c.c.  of  the  silver  nitrate  solution? 

17  mg.  per  cubic  centimetre  is  the  same  as  17 
grammes  per  litre;  therefore  the  silver  solution  is 
decinormal,  and  will  therefore  require  10  times 
its  volume  of  the  centinormal  potassium  chloride 
solution,  or  96.4  c.c. 


r 


CHAPTER  VII 


CALCULATION  OF  FORMULA 


The  organic  chemist  very  frequently  meets  the 
problem  of  determining  what  formula  to  give  a  sub- 
stance with  which  he  is  dealing.  In  order  to  do  this, 
he  must,  after  determining  what  the  constituents  of 
the  substance  are,  find  out  the  relative  quantities  of 
these  constituents.  This  really  amounts  to  finding 
the  percentage  composition  of  the  substance,  and 
though  it  is  not  actually  necessary  to  calculate  out 
the  percentage  composition,  this  is  usually  done,  and 
the  chemist  sets  down  the  figures  representing  the 
percentage  composition,  and  gives  the  formula  that 
he  calculates  from  these  figures. 

The  constituent  elements  of  the  substance  are  not, 
in  general,  separated  and  weighed,  but  they  are  con- 
verted into  compounds  whose  composition  is  already 
known.  For  instance,  it  is  not  feasible  to  separate 
the  carbon  from  alcohol  and  determine  its  amount 
in  that  way,  but  it  is  very  easy  to  burn  the  alcohol 

89 


I 


I 


IMJ 


I  I 

I! 


90 


ARITHMETIC  OF  CHEMISTRY 


il 


!:i 


I! 


Ill 


in  such  a  manner  that  all  the  carbon  dioxide  pro- 
duced may  be  collected  and  weighed.  In  the  same 
way,  hydrogen  is  weighed  as  water,  chlorine  is  con- 
verted into  silver  chloride,  and  sulphur  into  barium 
sulphate.  When  we  know  the  amount  of  each  of 
these  substances  produced,  there  is  no  difficulty  in 
calculating  the  amount  of  the  element  which  we  are 
considering. 

Carbon,  for  instance,  is  ^  the  weight  of  the  car< 
bon  dioxide  produced  by  its  combustion,  hydrogen 

is  A  the  weight  of  the  water,  chlorine  is    ■        the 
Iff  ^  143.5 

weight  of  the  silver  chloride  obtained,  and  sulphur 
2^  the  weight  of  the  barium  sulphate. 


Example.  —  When  i .  1 24  grammes  of  chloroform 
are  burned,  0.4139  gramme  of  carbon  dioxide  and 
0.0846  gramme  of  water  are  obtained.  If  the  same 
quantity  is  treated  in  the  proper  way,  it  yields  4.050 
grammes  of  silver  chloride.  These  are  all  of  the 
results  obtained  by  actual  analysis;  it  remains  for 
us  to  see  how  the  formula  of  chloroform  may  be 
deduced  from  these  data. 

We  know  that  when  12  grammes  of  carbon  are 
burned,  44  grammes  of  carbon  dioxide  are  obtained, 


r 


pro-    . 
same 
3  con- 
arium 
ich  of 
ilty  in 
we  are 

he  car- 
rdrogen 

il  the 
sulphur 

oroform 
ide  and 
lc  same 

IS  40SO 

of  the 

tins  for 

Imay  be 

Ibon  are 
Ibtained, 


CALCULATION  OF  FORMUL/E 


91 


and  therefore  that  the  amount  of  carbon  is  J|  of  the 
amount  of  carbon  dioxide  produced  by  its  combus- 
tion. The  amount  of  carbon  in  1.124  grammes  of 
chloroform  must  therefore  be 

0.4139  X  ^J  =  0.1 1 27  gramme. 

In  the  same  manner,  2  grammes  of  hydrogen  yield 
18  grammes  of  water,  and  therefore  the  amount  of 
hydrogen  is  ^g^  of  the  amount  of  water  produced  by 
the  combustion.  The  amount  of  hydrogen  in  1.124 
grammes  of  chloroform  is  therefore 

0.0846  X  ^  =  0.0094  gramme. 

Also  35.5  grammes  of  chlorine  gives  143.5  grammes 
of  silver  chloride,  and  therefore  the  amount  of  chlo- 
35.5 


I) 


rme  is 


143-5 


of  the  amount  of  silver  chloride.     The 


amount  of  chlorine  in  i .  1 24  grammes  of  chloroform 


is  therefore  4.050  x 


35-5 
143-5 


=  1.002  grammes. 


We  know  that  every  standard  quantity  of  carbon, 
which  we  represent  by  the  symbol  C,  weighs  1 2  times 
as  much  as  the  standard  quantity  of  hydrogen,  which 
we  represent  by  the  symbol  H  ;  so  that  if  the  weight 
of  carbon  actually  found  is  divided  by  12,  and  the 


r- 


92 


AR/TI/MKT/C  or  CHEMISTRY 


ii 


ti 


weight  of  the  hydrogen  by  unity,  and  the  quotient 
is  the  same,  we  know  that  there  arc  just  as  many 
standard  quantities  of  carbon  as  of  hydrogen.  If 
the  quotient  obtained  by  dividing  the  weight  of  car- 
bon by  12  is  one-half  as  great  as  that  obtained  by 
dividing  the  weight  of  hydrogen  by  unity,  we  know 
that  there  are  half  as  many  standard  quantities  of 
carbon  as  of  hydrogen.  If  the  quotient  is  twice  as 
great  in  the  first  case  as  in  the  second,  we  know 
that  there  are'  twice  as  many  standard  quantities  of 
carbon  as  of  hydrogen. 

In  the  same  way,  if  the  amount  of  chlorine  is 
divided  by  35.5,  the  quotient  obtained  will  show  how 
many  standard  quantities  of  chlorine,  represented  by 
the  symbol  CI,  there  are  for  every  standard  quantity 
of  carbon  or  of  hydrogen. 

In  the  case  of  the  numbers  we  have  found  for 
chloroform,  we  observe  that 


0.1 127 
12 


=  o.cx)94  =  0.0094  X  I 


0.0094 


0.0094  =  0.0094  X  I 


1.002 

35.5 


=  0.0282  =  0.0094  X  3 


for 


r 


CALCULATION  OF  FORMULAE 


93 


I'- 


3tient 
many 
L     If 
»f  car- 
ed by 
know 

ties  of 
vice  as 
\  know 
ities  of 

orine  is 
ow  how 
ted  by 
uantity 

md  for 


which  gives  the  ratio  of  the  standard  quantities  of 
carbon,  hydrogen,  and  chlorine.  The  formula  for 
chloroform  is  therefore  CHClg,  or  some  multiple  of 
it  such  as  C2H2CIQ,  CgHgClg,  etc. 

Simple  analysis  of  the  substance  does  not  decide 
between  these  formulae,  for  all  that  the  analysis 
proves  is  that  for  every  standard  quantity  of  carbon 
there  is  one  standard  quantity  of  hydrogen  and  three 
standard  quantities  of  chlorine. 

The  formula  which  expresses  in  the  simplest  form 
the  composition  of  a  substance  is  called  its  empirical 
formula,  and  this  may  or  may  not  be  the  formula 
that  best  represents  the  standard  quantity  of  the 
substance,  which  is  ordinarily  called  the  molecular 
formula. 

When  a  substance  is  volatile,  its  vapour  density 
is  of  very  great  importance  in  deciding  what  formula 
is  best  to  apply  to  it.  Other  considerations  are 
taken  into  account,  but  this  is  the  only  one  with 
which  we  are  immediately  concerned.  The  vapour 
density  of  chloroform  shows  that  its  correct  formula 
is  CHClg,  and  not  a  multiple. 

When  the  same  quantity  of  a  substance  is  used 
for    all    determinations,    we    have    seen     that    the 


94       ' 


ARITHMETIC  OF  CHEMISTRY 


empirical  formula  may  be  arrived  at  by  dividing 
the  quantity  of  each  element  by  the  standard  ivvight 
of  that  element,  the  quotients  obtained  showing  the 
relative  number  of  these  standard  weights.  In  fact, 
it  is  not  necessary  to  calculate  how  much  of  each 
element  is  present,  for  we  arrive  at  the  same  result 
if  we  divide  the  amount  of  carbon  dioxide  actually 
obtained  by  44  as  if  we  first  find  out  how  much  car- 
bon that  corresponds  to  and  divide  it  by  12.  So  also 
we  may  divide!  the  weight  of  water  by  9,  since  \ 
of  the  water  is  hydrogen ;  and  in  the  same  way  we 
may  divide  the  amount  of  silver  chloride  by  143.5. 

But  though  carbon  and  hydrogen  are  always  es- 
timated together,  chlorine,  sulphur,  and  nitrogen  are 
always  estimated  in  separate  experiments,  and  usu- 
ally with  different  amounts  of  substance.  1.124 
grammes  of  chloroform  are  far  too  much  to  take  for 
an  estimation  of  chlorine,  since  4.050  grammes  of 
silver  chloride  are  an  inconveniently  large  amount 
to  deal  with. 

But  though  different  amounts  of  substance  are 
taken  for  different  determinations,  it  is  easy  to  cal- 
culate from  the  data  obtained  how  much  of  each 
element  is   contained   in   one  gramme   of   the   sub- 


CALCULATION  OF  FORMULA 


95 


iding 
eight 
r  the 

fact, 

each  ^ 
result 
tually 
h  car- 
;o  also 
ince  \ 
ray  we 

I43-5- 
lys  es- 
en  are 
d  usu- 
1.124 
ke  for 
nes  of 
mount 

ce  are 
to  cal- 
each 
e  sub- 


stance, or  in  100  grammes;  in  other  words,  to  find 
the  percentage  composition.  The  results  of  analysis 
are  therefore  usually  worked  up  into  percentages 
and  put  down  in  that  form. 

Example. — 0.353   gramme  of  ethyl   sulphide  by 

combustion  gave  0.692  gramme   of  carbon  dioxide 

and  0.353   gramme  of  water.     Also,  0.241  gramme 

of  sulphide  gave  0.642  gramme  of  barium  sulphate. 

It  is  required  to  calculate  the  formula.     We  have 

Weight  of  carbon  dioxide  for  i  gramme 

=  0.692  X  ^^i^  =  1-959  grammes. 
Weight  of  water  for  i  gramme 

=  0.353  X  Ws^  =  1. 000  gramme. 
Weight  of  barium  sulphate  for  i  gramme 

=  0.624  X  -yyy^  =  2.583  grammes. 
Weight  of  carbon  for  i  gramme 

1.959  X  if   =0.5342=   5342% 
Weight  of  hydrogen  for  i  gramme 


i.ooo  X 


2     -_ 
15 


=  O.IIII  =    11.11% 


Weight  of  sulphur  for  i  gramme 

2.583  X  j^3%  =  0.3554  =  35-54% 

100.07 


I:. 


^ 


?'lll 

m 


96 


ARITHMETIC  OF  CHEMISTRY 


That  the  percentages  add  up  to  100.07  instead 
of  exactly  100  shows  a  slight  error  either  in  the 
estimation  of  the  constituents  or  in  the  calculations. 

To  get  the  formula  we  have 


53-42 
12 


Ratio 
4.45  =  I.II  X    4         C 


II. II 


=  ii.ii  =  I.II  X  10 


H 


35>54 
32 


=      I.II  =  I.II   X      I 


Therefore  the  formula  is  C4H1QS  or  some  multiple. 
As  a  matter  of  fact,  the  way  in  which  the  substance 
is  produced  indicates  that  the  above  is  the  correct 
formula. 

As  has  been  stated,  it  is  not  absolutely  necessary 
to  calculate  the  percentage  of  each  constituent  in 
the  substance,  though  the  data  must  always  be  such 
that  the  calculation  is  possible.  This  may  be  shown 
by  an  example,  and  we  shall  choose  one  in  which 
a  nitrogen  determination  is  involved. 

When  the  volume  (at  a  certain  temperature  and 
pressure)  of  nitrogen  obtained  from  a  quantity  of 
substance  is  irivcn,  it  is  easv  to  calculate  its  ^ 


easy 


ight 


CALCULATION  OF  FORMULA 


97 


istead 
n  the 
Ltions. 


altiple. 
stance 
orrect 

;essary 

lent  in 

such 

I  shown 

which 

|-e  and 
fity  of 
reight 


and  therefore  the  weight  which  would  be  obtained 
from  I  gramme  of  the  substance ;  and  the  remainder 
of  the  calculation  is  similar  to  the  preceding. 

Example.  —  It  is  required  to  calculate  the  formula 
of  ethylamine  from  the  following  data : 

I.  0.263  gramme  of  substance  gives  0.515  gramme 
of  carbon  dioxide  and  0.363  gramme  of  water. 

II.  0.328  gramme  produces  109.7  c.c.  of  nitrogen 
at  I5°C.  and  750  mm. 

The  weight  of    carbon   dioxide    produced    by    i 
gramme  is 

0-515  xJj<y>^=  1.955. 

The  weight  of   water  produced  by  i   gramme  is 

0.363x^^^=1.380. 
The  weight  of  nitrogen  produced  by  i  gramme  is 

1097  X  28  X  ^  X  752  X  I°50  =  o.3ii  gramme. 
22400  288     760      328 


1-955 
44 

1-380 
9 

0.311 
14 


Ratio 


=  0.0445  =  0.022  X  2 


=     0.153  =0.022  X  7 


=     0.022  =  0.022  X   I 


H 


N 


.  '• 


u 


98 


AKfTHMJiT/C  OF  CHHM/SJKV 


I     ' 

i' 


iii 


11! 


Therefore  the  formula  is  CjH^N. 

The  results  obtained  in  actual  practice  are  not 
so  exact  as  the  figures  given  above.  The  estima- 
tion of  carbon  is  usually  too  low,  and  the  estimation 
of  hydrogen  too  high.  The  formula  is  chosen  which 
most  nearly  represents  the  results  of  experiment 
and  suits  the  chemical  character  of  the  substance. 
Then  the  percentages  of  the  elements  are  put  down 
as  calculated  from  this  formula  and  side  by  side, 
for  comparison,  the  percentages  actually  obtained. 

Example.  —  I.  0.1520  gramme  of  a  substance 
gave  0.3352  gramme  of  carbon  dioxide  and  0.0708 
gramme  of  water. 

II  0.2071  gramme  of  the  substance  gave  64.5 
c.c.  of  nitrogen  at  23^.5  C.  and  774.5  mm. 


Calculated  from 

formula  CgHgNi 

Found 

c 

60 

60.16 

H 

5 

5.18 

N 

35 

35.60 

100 

I00.Q4 

So  that  the  formula  was  probably  C8H8N4,  though 
the  analysis  was  not  very  satisfactory.     The  follow- 


64.5 


CALCULATION  OF  FORMULAE 


99 


ing   results  are  more   satisfactory.     An  analysis  of 
parabromphenyl  hydrazin  gave  these  data: 

I.  0.234  gramme  of  substance  gave  0.328  gramme 
of  carbon  dioxide  and  0.082  gramme  of  water. 

II.  0.132    gramme    gave    17.5    c.c.   of    nitrogen 
measured  at  i7°C.  and  742  mm. 

III.  0.1435  gramme  gave  0.1435  gramme  of  silver 
bromide. 


Calculated  from  formula 

1 

CoIlTNoHr 

Found 

c 

38.50 

38.28 

H 

371 

3.93 

N 

14.97 

14.94 

Br 

42.72 

42.56 

100.00 


99.71 


These  results  are  better  than  those  in  the  last 
case,  not  only  because  the  percentages  found  add 
up  to  a  number  more  nearly  equal  to  100,  but 
because  the  error  in  each  determination  is  in  the 
direction  which  is  to  be  naturally  expected. 

Sometimes  two  formulas  are  given,  and  the  per- 
centages calculated  from  them  are  put  side  by  side 
with  those  actually  obtained  by  experiment. 


V 


% 


100 


ARITHMETIC  OF  CHEMISTRY 


Example.  —  I.    0.1541   gramme  of    a    substance  ^ 
gave  0.4426  gramme  of  carbon  dioxide  and  0.1128 
gramme  of  water. 

II.  0.1097  gramme  gave  0.3164  gramme  of  carbon 
dioxide  and  0.0824  gramme  of  water. 

III.  0.3014  gramme  gave   36.1    c.c.  of  nitrogen 
at  21°  C.  and  743.5  mm. 


m 


Calculated  for  formula 

Found 

C7II9N 

.   QHtN 

I 

II 

c 

78.50 

77.42 

78.22 

78.58 

H 

8.41 

7-53 

8.13 

8.35 

N 

13.09 

15.06 

III 


13.0 

Here  the  first  formula  is  the  better  one,  as  is 
especially  seen  in  the  nitrogen  determination. 

It  must  always  be  borne  in  mind  that  the  results 
of  analysis  do  not  decide  between  a  formula  and  any 
multiple  of  that  formula.  This  matter  can  be  de- 
cided only  by  other  considerations,  and  among  these 
the  vapour  density  (for  any  substance  that  will  vola- 
tilize) is  a  very  important  one. 

Somewhat  similar  calculations  are  made  for  inor- 
ganic substances.  In  these  cases  it  is  sometimes 
required  to  calculate  the  standard  (or  atomic)  weight 


CALCULATION  OF  FORMULA 


lOI 


stance 
D.1128 

:arbon 

trogen 


III 


130 
as  is 

results 
id  any 
be  de- 
these 
1  vola- 

r  inor- 
letimes 
weight 


of  a  metal  as  well  as  the  formula  of  the  compound. 
An  important  point  in  such  determination  is  the  spe- 
cific heat  of  the  element,  which,  multiplied  by  the 
standard  weight,  gives  a  product  approximately  equal 
to  six.  The  vapour  density  of  compounds  also  is  of 
importance  in  this  connection. 

Problems  are  sometimes  set  in  the  calculation  of 
the  formulae  of  minerals,  in  which  the  percentage  of 
each  element  is  given.  These  problems  may  be 
worked  in  exactly  the  same  way  as  when  the  formula 
of  an  organic  compound  is  required. 

In  minerals,  however,  the  constituents  vary  largely, 
one  metal  being  replaced  in  part  by  another  which 
gives  isomorphous  compounds.  In  the  case  of  min- 
erals containing  oxygen,  such  as  carbonates  and  sili- 
cates, the  metal  is  usually  estimated  as  oxide,  and  the 
percentages  given  in  that  way. 

The  following  is  an  example  given  in  the  ordinary 
form: 


CaO 

28.4 

MgO 

12.3 

FeO 

12.3 

MnO 

1.9 

CO2 

44.4 

99-3 

.  i.-l 


102 


ARITHMETIC  OF  CHEMISTRY 


That  the  numbers  do  not  add  up  to  exactly  icxD  is 
due  to  the  difficulty  in  making  a  perfectly  accurate 
analysis. 

In  the  same  manner  as  the  ratios  between  the  ele^ 
mcnts  of  organic  compounds  were  calculated,  we  may 
calculate  the  ratios  of  the  oxides.  The  quantity  of 
calcium  oxide  must  be  divided  by  56,  of  magnesium 
oxide  by  40,  and  so  on. 

Doing  this  with  the  mineral  in  question,  we  have : 

Ratio 
28  A. 

Calcium  oxide,        28.4% ;  — ^  =  0.507 

Magnesium  oxide,  12.3%;  — ^  =  0.307 

40 

Ferrous  oxide,        12.3%;  — —  =  o.  1 70  [  0.504 

72 

Manganese  oxide,    1.9%;    —=0.027 

Carbon  dioxide,      44.4%;  ^^=  1.009 

44 

The  ratio  of  the  calcium  oxide  to  the  magnesium 
oxide,  along  with  the  oxides  which  replace  it  isomor- 
phously,  is  I  :  I,  and  to  the  carbon  dioxide  i  :  2. 

Therefore  the  formula  may  be  written 

CaO .  (Mg,  Fe,  Mn)0 . 2  COg 


lesium 
iomor- 


CALCULATION  OF  FORMULA 


103 


or  Ca(Mg,  Fe,  Mn)(C03)2;  or  it  may  be  considered  as 
a  mixture  of  the  isomorphous  substances,  CaMg(C0g)2, 
CaFe(C03)2,  and  CaMn(C03)2,  the  relative  amounts 
of  these  substances  being  shown  by  the  ratios  307, 
170,  and  27.  If  the  mineral  were  pure  dolomite,  we 
should  have  more  magnesium  oxide,  instead  of  the 
iron  oxide  and  manganese  oxide.  It  is  interesting  to 
calculate  the  amount  of  magnesium  oxide  corres'pond- 
ing  to  those  other  oxides,  and  thus  obtain  the  per- 
centage composition  of  pure  dolomite  from  the  re- 
sults of  this  analysis. 

The  weight  of  MgO  is  evidently  f|  that  of  FeO 
and  ^\  that  of  MnO.  In  the  ratios  obtained  above 
we  have  already  divided  by  the  denominators  of 
these  fractions;  hence,  to  obtain  the  amounts  of 
MgO,  corresponding  to  the  FeO  and  MnO,  we 
merely  need  to  multiply  the  ratios  by  40. 

Doing  this  we  have  : 

Ratio 

FeO   0.170  X  40=    6.8%  MgO  equivalent  to  FeO. 
MnO  0.027  X  40  =  I.I  %  MgO  equivalent  to  MnO. 

12.3%  MgO  in  analysis. 

20.2%  Total  MgO. 


/ 1 


'% 


104 


ARITHMETIC  OF  CHEMISTRY 


This  amount  of  magnesium  oxide  added  to  the 

calcium  oxide  and  carbon  dioxide  does  not  amount 

to  100,  but  to  93 ;  but  we  may  easily  calculate  it  into 

percentages. 

CaO     28.4-!- 0.93=    30.5% 

MgO    20.2  +  0.93=   21.7 

COa      44.4  -*-  0.93  =   47.8 

93.  100.  % 

and  this  percentage  corresponds  with  the  formula 
CaO.  MgO.  2  CO2  or  CaMg  (COg)^. 

A  sample  of  pyroxene  may  be  taken  as  another 
illustration. 

55.40%; 


Silica 

Ferrous  oxide  2.50%; 

M  anganese  oxide      2.83%; 


55.40 


60 
2.50 

2^ 

71 


=0.923 
=0.035 

=  0.040 


22.57 

Magnesium  oxide    22.57%;    — ^^^=0.564 

40 


Calcium  oxide  15.70%; 


1570 
56 


=0.280 


0.919 


Ratio,  1:1;  Formula,  (Mg,  Ca,  Mn,  Fc)0,  SiOg. 

In  order  to  show  what  variations  may  occur,  and 
how  nearly  the  ratios  approach  to  the  ideal,  the 
analyses  of  different  garnets  may  be  given. 


)  the 
lount 
t  into 


)rmula 


nother 


0.919 


ir,  and 
il,   the 


CALCULATION  OF  I  OKMUL.E 


105 


Garnets. 


SiOa 

AlaOg 

FePa 

CraOg 

FeO 

MnO 

MgO 

CaO 


I 
38.25% 
19.35 
733 


II 

37.n% 
5.88 


III 

36.65% 
17.50 


0.50 
2.40 

3 '.75 
99.58% 


22.54 
2.44 

1. 10 

3034 
99.41% 


6.20 
4.97 

0.81 
33-20 


99.3370 


From      I.  SiOa  :  (AI2O3,  FcaOg) :  (MnO,  MgO,  CaO) 
=  0.637:0.235:0.634; 

From    II.  SiO.^ :  (AI2O3,  CrgOg) :  (FeO,  MgO,  CaO) 
=  0.618  : 0.198  : 0.602; 

From  III.  Si02:(Al208,  Cr203):(FeO,  MgO,  CaO) 
=  0.611  : 0.214  : 0.681  ; 

therefore  the  probable  formula  for  garnet  is 

3  RO  •  R2O8  •  3  Si02. 

Though  calculations  do  not  depend  upon  any 
theory  regarding  the  constitution  of  matter,  all  the 
facts  described  find  a  ready  explanation  if  the  exist- 
ence of  atoms  is  admitted,  matter  being  considered 
as  composed  of  indivisible  particles.     In  this  way  it 


ii 


K  'V    ! 


To6 


ARITHMETIC  OF  CHEM/STRV 


is  seen  why  the  quantity  of  oxygen  united  to  a 
given  quantity  of  carbon  should  be  exactly  twice  as 
much  in  carbon  dioxide  as  in  carbon  monoxide. 

The  kinelic  theory  of  gases  gives  support  to  the 
idea  of  molecules  and  to  Avogadro's  law.  It  is  not 
the  purpose  of  this  little  book,  however,  to  deal  with 
the  theoretical  side  of  the  subject. 

EXAMPLES 

1.  State  concisely  how  the  combining  weight  of  an 
element,  as,  for  example,  nitrogen,  can  be  found  from 
vapour  density  determinations  and  analyses.  (Cornell 
Univ.,  March,  1898.) 

Analysis  of  ammonia  shows  the  volume  of  the  contained 
hydrogen  to  be  3  times  that  of  the  nitrogen,  and  this  with 
the  vapour  density  determines  the  formula  NH3.  The 
weight  of  the  nitrogen  is  4I  that  of  the  hydrogen,  therefore 
the  weight  represented  by  the  symbol  N  is  14  times  that 
represented  by  the  symbol  H. 

2.  Alcohol  contains 

Carbon 52.17% 

Oxygen 34.79 

Hydrogen   .......     13.04 

The  specific  gravity  of  its  vapour  is  1.593  (referred  to  air)  ; 
find  its  molecular  weight.     (Cornell  Univ.,  June,  1896.) 

Ans.  CgHflO. 


to  a 
ice  as 

ie. 

;o  the 
is  not 
.1  with 


of  an 
i  from 
Cornell 

ntained 
tiis  with 
The 
lerefore 
les  that 


to  air)  ; 
96.) 
C2H6O. 


CALCULATION  OF  FORMULAE  I07 

3.  One  gramme  of  acetic  acid  gave  on  combustion 
1.466  grammes  of  carbon  dioxide  and  0.600  of  water. 
The  vapour  density  of  acetic  acid  is  30.  Calculate  the 
percentage  of  carbon,  hydrogen,  and  oxygen ;  give  the 
empirical  and  molecular  formula.  (Columbia  Univ.,  May, 
1898.)  Ans.  Carbon  =  40%,  hydrogen  =  6.66%. 

The  percentage  of  oxygen  is  obtained  by  subtracting  the 
sum  of  the  carbon  and  hydrogen  from  100,  and  is  53.34%. 
Empirical  formula  CH^O,  but  since  this  formula  represents 
22.4  litres  of  vapour  as  weighing  only  30  grammes,  that  is, 
only  fifteen  times  as  much  as  hydrogen,  the  molecular 
formula  must  be  C^H^O^,  which  is  thirty  times  as  heavy 
as  hydrogen. 

4.  One  gramme  of  alcohol  gave,  on  combustion,  carbon 
dioxide  1.913,  water  1.173.  The  vapour  density  of  alcohol 
is  23 ;  calculate  the  percentage  of  carbon,  hydrogen,  and 
oxygen ;  give  the  empirical  and  molecular  formulae.  (Co- 
lumbia Univ.,  May,  1898.) 

Ans.     Carbon 52.2 

Hydrogen 13.0 

Oxygen 34.8 

Empirical  formula C2H0O 

Molecular  formula CaHgO 

5.  Calculate  the  formula  of  mellitic  acid  from  the  fol- 
lowing data : 

Carbon 42.1% 

Hydrogen i-7 

Oxygen 56.2 

1 00.0 


I 


I08 


AJarJ/MhT/C  OF  CHEMISTRY 


■: 


'1 


f 


;'  ■ 


u 

)'\  'I 


All  the  hydrogen  can  be  replaced  by  a  metal.  It  is 
hexabasic.     (Harvard  Univ.,  1885.) 

The  simplest  or  empirical  formula  is  C^HOo,  but  since 
it  is  hexabasic  and  all  the  hydrogens  may  be  replaced  by 
metal,  the  formula  as  above  given  must  be  multiplied  by 
6;   therefore  the  molecular  formula  is  CiaH„0|2. 

6.  Calculate  the  atomic  weight  of  molybdenum  from 
the  following  data : 

Percentage  cumpositiun  of  chloride 

Molybdenum 40'33 

Chlorine 59-67 

100.00 

Specific  heat  of  molybdenum  =  0.064.  (Harvard  Univ., 
1887.) 

Since  the  specific  heat  multiplied  by  the  atomic  weight 
should  give  a  number  approximately  equal  to  6,  the  atomic 
weight  of  molybdenum  should  be  in  the  neighbourhood 
of  94. 

But  the  quantity  of  the  chlorine  is  approximately  f  as 
great  as  the  molybdenum,  therefore  94  of  molybdenum 
would  need  141  of  chlorine.  But  141  is  approximately 
4  ^  35  "5  >  therefore  we  may  assume  that  the  chloride  in 
question  has  the  formula  M0CI4. 

With  this  assumption,  we  calculate  from  the  data  — 


9. 

lowii 


Atomic  weight  of  molybdenum  =  40.33  x 


142 
59^67 


=  96 


(S; 


CAIXULATtON  OF  I'ORMUL^ 


109 


It   is 

since 
ed  by 
ed  by 

I  from 


Univ., 

;  weight 

atomic 

)urhood 

ly  \  as 
bdenum 
cimately 
aride  in 


=  96 


7.  Determine  the  atomic  weight  of  glucinum  from  the 
following  data : 

The  vapour  density  of  glucinic  chloride  is  40.2,  and  it 
contains  11.7%  of  glucinum.     (Harvard  Univ.,  1885.) 

The  vapour  density  being  40.2,  the  standard  (or  molec- 
ular) weight  is  80.4;  and  the  chlorine  must  be  71,  there- 
fore the  glucinum  is  9.4  (or  some  submultiple).  This 
agrees  exactly  with  the  analysis  given.  That  9.4  is  more 
likely  correct  than  any  submultiple  is  proved  by  other  con- 
siderations. 

8.  A  metal  X  forms  a  volatile  chloride  containing 

A' =34.46 
Chlorine  =  65.54 

The  vapour  density  of  the  chloride  referred  to  hydrogen 
is  81.25.  1^®  specific  heat  of  the  metal  is  0.11379.  Cal- 
culate the  atomic  weight  of  X  and  give  the  formula  of  the 
chloride.     (London  Univ.  Int.  Sci.,  1893.) 

From  the  specific  heat,  the  atomic  weight  of  X  is  between 
50  and  60,  and  therefore  from  the  vapour  density  and  a 
superficial  glance  at  the  proportion  of  the  elements  in  the 
chloride,  it  appears  that  the  formula  is  XCI3.  Then  making 
use  of  the  exact  figures,  the  atomic  weight  of  X  is  found  to 
be  56. 

9.  Calculate  the  formula  of  the  substance  from  the  fol- 
lowing percentage  composition : 

Silicon  26.27  Calcium  18.43 

Magnesium  11.06  Oxygen  44.24 

(Sydney  Univ.,  1894.)  Ans.  CaMgSi20fl. 


no 


Al<irU.\tiaiL •  01'    (  UEMISTR 1  • 


: 


i 


! 


10.  Two  specimens  of  chromile  upon  analysis  gave  the 
following  iH'nentagcs  : 

I  11 

55''4% 

5-75 
1  2.o6 

18.02 

9-39 
100.36 

Atis.  KO  .  RA. 

11.  What  is  the  formula  of  the  mineial  three  samples  of 
which  gave  on  analysis  : 

I  II 

i7.«3% 
28.79 

100.00 


Cr,(), 

55.37% 

AM)... 

>3-97 

!•>,(),, 

1.10 

FeO 

1  H.04 

MgO 

10.04 

What  is 

the  typic 

al  formula? 

S        15-92% 
Ag      52.71 
C'u       30.95 

99.58 


III 

19.93% 
24.04 

53-94 
97.91 

A  us.   (Ag,  Cu)aS. 


12.   Two  sam])les  of  erythtite,  on  analysis,  gave  the  fol- 
lowing percentages.     What  is  the  formula  of  the  mineral? 

I  11 

36.42% 

23.75 
11.26 

3.51 
0.42 

23.52 

■ins.  3  RC)  .  AsA  •  8  H^O. 


AsA 

3>^-43% 

CoO 

36.52 

NiO 

FeO 

l.OI 

CaO 

HaO 

24.10 

II 


<l  il 


vc  llic 


pics  of 


he  fol- 
leral  ? 


8  H,0. 


CALCl^I. AT/OX  or  l'ONMUI..K 


I  I  I 


13.    IVnfu'ld  found  hy  analysis  of  Iriphylitc  the  following 
composition.     What  is  the  forninia  of  the  mineral? 


FcO 

MnO 

C:a() 

MgO 

Li,() 

NiiaO 


44.70 

26.40 

17.84 

0.24 

0.47 

936 
0.35 


Ans,  Vl\  ■  2(l<'e,  Mn,  Ca,  Mg)0  •  (Lij,  Na,)©,  or 
chielly  Li(Ke,  lVln)l»04. 

Miscellaneous  Examples 

1.  What  volume  of  chlorine  should  be  obtained  at  20°  C. 
and  760  nnu.,  by  acting  on  i  gramme  of  manganese  dioxide 
with  hydrochloric  acid?     (Univ.  of  Sydney,  1H95.) 

Ans.    276.3  CO. 

2.  What  volume  of  carbon  monoxide,  at  a  temperature 
819"  ('.  and  i)ressure  720  mm.,  would  be  rerpiired  to  reduce 
1  kilogramme  of  ferrous  oxide  to  metallic  iron?  (Univ. 
of  Sydney,  1895.)  Ans,    1313.5  litres. 

3.  How  many  litres  of  oxygen  are  rcfiuircd  for  the 
complete  combustion  of  2  litres  of  each  of  the  follow- 
ing :  {a)  carbon  monoxide,  (/^)  methane,  {c)  ethylene, 
{ji)    acetylene?     (Univ.  of  Sydney,  1896.) 

Ans.    i^a)  I  litre ;  {J))  4  litres ;  {c)  G  litres ;  (^/)  5  litres. 


112 


ARITHMETIC  OF  CHEMISTRY 


l:!i 


I  >  ■Hill 


i:  '\ 


■J 

I 


4.  A  monobasic  organic  acid  has  the  empirical  formula 
CHoO ;  0.296  gramme  of  its  silver  salt  gave  on  ignition 
0.162  gramme  of  silver.  What  is  its  molecular  weight 
and  molecular  formula?     (Univ.  of  Sydney,  1896.) 

Since  the  acid  is  monobasic,  the  silver  salt  has  the 
formula  AgX,  where  X  stands  for  all  that  is  not  silver,  and 
the  acid  itself  has  the  formula  HX.  The  molecular  weight 
of  the  silver  salt  is  therefore  108  x  y||  =  197;  hence  the 
molecular  weight  of  the  acid  HX  is  90,  which  is  three  times 
the  weight  represented  by  the  empirical  formula  CH2O. 
The  molecular  formula,  therefore,  of  the  acid  is  C3H6O3. 

5.  What  is  the  result  of  the  action  of  chlorine  on  am- 
monia? How  much  chlorine  would  be  required  to  decom- 
pose 1.7  grains  of  ammonia?  (McGill  Univ.  Med.  Exam., 
1897.)  Ans.   2.66  grains. 

6.  The  composition  of  a  compound  is 

Barium 46.12 

Sulphur 21.54 

Oxygen 32.32 

99.98 

What  is  the  formula  and  name  of  the  compound?     (First 
Professional  Exam.,  Glasgow  Univ.,  1888.) 

Ans.   Barium  dithionate ;  BaS206c 

7.  A  substance  has  the  following  composition :  carbon 
10.04%,  hydrogen  0.83%,  chlorine  89.13%,  and  its  vapour 
density  is  59.75.  What  is  its  name  and  formula?  (Fii^t 
Professional  Exain.,  Glasgow,  1888.) 

Ans.   Chloroform ;  CHCI3. 


(First 


;hci,. 


CALCULATlOISf  OF  FORMULAC  II3 

8.  A  litre  of  a  mineral  water  yielded  0.0134  gramme  of 
silver  iodide.  How  much  iodine  is  contained  in  100,000 
c.c.  of  the  water?  (First  Professional  Exam.,  Glasgow 
Univ.,  1889.)  Ans.  0.7241  gramme. 

9.  How  many  litres  of  oxygen,  at  12°  C.  and  750  mm. 
pressure,  can  be  got  by  heating  100  grammes  of  manga- 
nese dioxide?     (First  Professional  Exam.,  Glasgow  Univ., 

1890.) 

3  MnOo  =  Mn;.04  +  Oj.     Ans.  9.076  litres. 

10.  Calculate  the  weight  of  zinc  and  hydrochloric  acid, 
respectively,  which  would  be  required  to  produce  hydrogen 
to  inflate  a  balloon  of  1000  cubic  metres  capacity,  at  a  tem- 
perature of  15°  C,  and  pressure  of  760  mm.  What  would 
be  the  bulk  of  the  gas  when  the  barometer  had  fallen  to 
730  mm.,  and  the  temperature  was  —  6°  C.?  (First  Pro- 
fessional  Exam.,  Glasgow  Univ.,  1891.) 

I  cubic  metre  =  1000  litres. 
Ans.    Zinc,    2771.8    kilogrammes;     hydrochloric    acid, 
3089.2  kilogrammes;  965.18  cubic  metres. 

11.  What  are  the  formula  and  name  of  the  salt  having  the 
following  percentage  composition  ? 

Calcium 38.72 

Phosphorus 20.00 

Oxygen 41.28 

(First  Professional  Exam.,  Glasgow  Univ.,  1892.) 

Ans.   Tricalcic  phosphate  ;  Ca;i(P04)2. 

12.  A  liquid  containing  i  gramme  of  sodium  chloride 
was  added  to  another  containing  2  grammes  of  silver  ni- 

I 


'III 


114 


ARITHMETIC  OF  CHEMISTRY 


trate.  Whether  did  the  filtered  liquid  contain  sodium 
chloride  or  silver  nitrate,  and  how  much?  (First  Profes- 
sional Exam.,  Glasgow  Univ.,  1892.) 

Ans.   Sodium  chloride ;  0.3 1 1 7  gramme. 

13.  What  volume  of  hydrogen  and  oxygen,  respectively, 
could  be  got  from  a  cubic  centimetre  of  water  at  its  maxi- 
mum density  point?  (First  Professional  Exam.,  Glasgow 
Univ.,  1892.) 

Ans,   Hydrogen,  1.244  litres  ;  oxygen,  0.622  litre. 

14.  To  what  temperature  centigrade  would  30  litres  of 

hydrogen  measured  at    70°  F.  require  to   be   reduced  to 

occupy   20   litres?      (First    Professional   Exam.,   Glasgow 

Univ.,  1893.) 

70°  F.=  21.1°  C.      Ans.  -  76.93°  C. 

15.  Calculate  what  volume  of  nitrogen,  measured  at 
15°  C.  and  750  mm.,  could  be  obtained  from  i  gramme 
of  urea  by  the  action  of  sodium  hypobromite,  assuming  all 
the  nitrogen  to  be  liberated  in  the  process.  (First  Pro- 
fessional Exam.,  Glasgow  Univ.,  1896.) 

It  is  only  necessary  for  this  problem  to  know  the  formula 
of  urea,  since  we  are  told  that  all  the  nitrogen  is  liberated. 
Urea  has  the  formula  CO(NH2)2.  Ans,   251.71  c.c. 

.  16.  What  weights  of  ammonium  chloride  and  quick  lime 
must  be  taken  in  order  to  produce  2  litres  of  ammonia  — 
the  gas  to  be  measured  at  17°  C.  and  750  mm.?  (First 
Professional  Exam.,  Glasgow  Univ.,  1897.) 

Ans.  Ammonium  chloride,  4. 439  grammes ;  quick  lime, 
2.963  grammes. 


^iili 


CALCULATION  OF  FORMULA 


115 


I 


sodium 
Profes- 

:amme. 

ctively, 
)  maxi- 
ilasgow 

!2  litre. 

itres  of 
iced  to 
Glasgow 

3.93°  C. 

ired  at 
jramme 
Cling  all 
St  Pro- 

brmula 
)erated. 
71  c.c. 

ck  lime 
onia  — 
(First 

k  lime, 


17.  What  volume  of  oxygen  gas,  measured  at  700  mm. 
pressure  and  10°  C.  temperature,  can  be  obtained  from  100 
grammes  of  potassium  chlorate?  (First  Professional  Exam., 
Edin.  Univ.,  1888.)  Ans.   30.87  litres. 

18.  How  many  litres  of  oxygen,  measured  at  15°  C.  and 
750  mm.,  are  required  for  the  complete  combustion  of  100 
grammes  of  marsh  gas?  (First  Professional  I'^xam.,  Edin. 
Univ.,  1891.)  Ans.    299.3  litres. 

19.  How  many  cubic  centimetres  of  carbonic  anhydride, 
measured  at  730  mm.  and  20°  C,  would  be  obtained  by 
the  complete  combustion  of  i  gramme  of  cane  sugar? 
(First  Professional  Exam.,  Edin.  Univ.,  1891.) 

Ans.   878.4  c.c. 

20.  What  would  be  the  weight  of  the  products  of  the 
complete  combustion  of  500  grammes  of  ethylene,  and 
what  would  be  the  volume  in  litres  of  the  gaseous  product 
measured  at  10°  C.  and  740  mm.?  (First  Professional 
Exam.,  Edin.  Univ.,  1892.) 

Ans.    2214.3  grammes ;  851.73  litres. 

21.  Hydrochloric  acid  is  diluted  with  water  until  100  c.c. 
of  the  dilute  solution  exactly  precipitate  0.34  gramme  of 
silver  nitrate;  how  many  cubic  centimetres  of  this  dilute 
hydrochloric  acid  are  required  exactly  to  neutralize  0.68 
gramme  of  ammonia?  (First  Professional  Exam.,  Edin. 
Univ.,  March,  1893.)  Ans.   2000  c.c. 

22.  What  percentage  of  iron  is  contained  in  pure  ferrous 
carbonate?  (First  Professional  Exam.,  Edin.  Univ.,  July, 
1893.)  Ans.  48.29%. 


1,1 


ii6 


ARITHMETIC  OF  CHEMISTRY 


!>'■   ''I 


III 


1.1 


i;  ■ 


23.  What  volume  in  litres  of  carbonic  oxide,  at  760  mm. 
and  0°  C,  can  be  obtained  from  10  grammes  of  sodium 
formate,  by  the  action  of  strong  sulphuric  acid?  (First 
Professional  Exam.,  Edin.  Univ.,  1894.) 

The  equation  representing  the  reaction  is 

2  HCOONa  +  H2SO4  =  Na2S04  +  2  H,0  +  2  CO 

Ans.   3.294  litres. 

24.  What  volume  of  nitrogen  at  standard  temperature 
and  pressure  can  be  obtained  from  100  grammes  of  ammo- 
nium nitrite?  (First  Professional  Exam.,  Edin.  Univ., 
1894.)  I  Ans.   35  litres. 

25.  50  c.c.  of  a  given  solution  of  hydrochloric  acid  are 
exactly  sufficient  to  precipitate  as  silver  chloride  the  whole 
of  the  silver  from  2  grammes  of  silver  nitrate.  What  weight 
of  caustic  soda  would  be  exactly  neutralized  by  100  c.c.  of 
this  solution  ?  Calculate  the  result  to  three  places  of  deci- 
mals.    (First  Professional  Exam.,  Edin.  Univ.,  July,  1894.) 

Ans.   0.941  gramme. 

26.  Calculate  to  two  places  of  decimals  the  percentage 
composition  of  silver. acetate.  (First  Professional  Exam., 
Edin.  Univ.,  July,  1895.) 

Ans.  Silver,  64.67%  ;  carbon,  14.37% ;  hydrogen, 
1.79%  ;  oxygen,  19.16%. 

27.  What  weight  of  metallic  zinc  is  required  for  the 
precipitation  of  5  grammes  of  metallic  silver  from  a  solu- 
tion of  silver  acetate?  (First  Professional  Exam.,  Edin. 
Univ.,  March,  1896.)  Ans.    1.470  grammes. 


CALCULATION  OF  FORMULA 


117 


28.  Calculate  the  percentage  composition  of  urea,,  to 
two  places  of  decimals.  (First  Professional  Exam.,  Edin. 
Univ.,  1896.) 

Ans.  Carbon,  20.00% ;  oxygen,  26.67%  \  nitrogen, 
46.67%  ;  hydrogen,  6.66%. 

29.  What  weight  in  grammes  of  silver  oxide  must  be 
taken  to  obtain,  by  heating  it,  i  litre  of  oxygen  gas  meas- 
ured at  0°  C.  and  760  mm.?  (First  Professional  Exam., 
Edin.  Univ.,  July,  1897.)  Ans.   20.72  grammes. 

30.  What  volume  in  litres  of  hydrogen,  measured  at  0°  C. 
and  760  mm.  pressure,  would  be  produced  by  the  action  of 
20  grammes  of  sodium  on  water?  What  would  be  the 
volume  at  18°  C.  and  750  mm.  ?  (First  Professional  Exam., 
Edin.  Univ.,  March,  1898.)    Ans.  9. 739 litres;  10.517  litres. 


APPENDICES 


1 

a  lil 

hun 

cent 

for  t 

pies 

meti 

T 

equa 

cent! 

it  is 

of  w 


A 

weigh 
divide 
The  V 
I  kilo 


APPENDIX   A 


THE  FRENCH   SYSTEM   OF  MEASURES 

The  French  standard  of  length  is  the  metre,  which  is 
a  little  more  than  39.37  inches.  It  is  divided  into  tenths, 
hundredths,  and  thousandths,  called  respectively  decimetre, 
centimetre,  and  millimetre ;  the  Latin  prefixes  being  used 
for  the  fractions.  Greek  prefixes  are  used  to  denote  multi- 
ples of  the  standard,  the  most  important  being  the  kilo- 
metre, which  is  1000  metres  or  0.6214  mile. 

The  standard  capacity  is  the  litre  or  cubic  decimetre, 
equal  to  1.76  pints.  It  is  also  subdivided  into  decilitre, 
centilitre,  and  millilitre,  but,  instead  of  using  these  terms, 
it  is  more  common  to  use  the  term  cubic  centimetre,  1000 
of  which  make  a  litre. 

10  millilitres    or       10  c.c.  =  i  centilitre 
10  ceniilitres  or     100  c.c.  =  i  decilitre 
10  decilitres    or  1000  c.c.  =  i  litre 

A  cubic  centimetre  of  water  at  the  temperature  of  4°  C. 

weighs  a  gramme,  which  is  the  standard  of  iveight,  and  is 

divided  into  decigrammes,  centigrammes,  and  milligrammes. 

The  weight  of  a  litre  of  water  at  4°C.  is  1000  grammes  or 

I  kilogramme,  equal  nearly  to  2.2  pounds  avoirdupois. 

121 


122 


ARITHMETIC  OF  CHEMISrRY 


APPENDIX   H 


ARITHMETICAL  CALCULATIONS 


The  full  form  of  the  calculation  on  page  19  is 

96  grammes  of  oxygen  are  prepared  from  245  grammes  of 
potassium  chlorate. 

I  gramme  of  oxygen  is  prepared  from  245  x  ^V  gramme 
of  potassium  chlorate. 

80  grammes  of  oxygen  are  prepared  from  245  x  §§  gramme 
of  potassium  chlorate. 

It  should  never  be  necessary  for  the  student  to  go  through 
all  of  the  steps,  however.  To  i)repare  80  grammes  of  oxy- 
gen evidently  requires  less  chlorate  than  to  prepare  96 
grammes ;  and  it  is  almost  equally  evident  that  it  requires 
1^  as  much.  The  student  can  always  be  kept  right  in  the 
manipulation  of  the  fractions  by  asking  the  simple  question 
whether  the  answer  should  be  more  or  less  than  the  number 
given  (by  the  equation  or  otherwise),  and  then  by  arrang- 
ing the  terms  of  the  fraction  so  as  to  obtain  the  required 
result.  This  is  really  the  old  Rule  of  Three  expressed  in 
different  form,  and  I  should  not  have  thought  it  necessary  to 
write  this  appendix,  except  that  I  have  found  by  experience 
that  blunders  are  made  by  students  from  whom  one  might 
least  expect  it.  The  method  of  solving  problems  from  the 
common-sense  point  of  view  rather  than  by  direct  applica- 
tion of  formula  has  been  frequently  followed  in  the  text, 
and  is  the  one  which  is  best  fitted  to  prevent  slips  in  the 
arithmetic. 


APPEXniCES 


123 


APPENDIX   C 

COMPARISON  OF  TIIKKMOMKTRIC  SCALES 

Thf.rk  are  two  thermometric  scales  in  wliich  the  melting 
point  of  ice  is  taken  as  the  zero.  These  are  the  Centigrade 
and  R<Jaumur  scales.  In  the  former  the  boiling  point  of 
water  is  called  100  (whence  the  name  centigrade);  in  the 
latter,  80. 

In  the  Fahrenheit  scale  the  distance  between  the  melting 
point  of  ice  and  the  boiling  point  of  water  is  divided  into 
180  parts;  but  the  melting  of  ice  is  not  the  zero,  but  is 
called  32°,  and  the  zero  probably  marks  what  Fahrenheit 
considered  the  lowest  attainable  teni[)erature. 

The  method  of  changing  from  one  scale  to  the  other  is 
so  clearly  given  in  Tait's  "  Heat  "  that  I  transcribe  it :  "If 
we  suppose  the  same  thermometer  to  have  these  three  sepa- 
rate scales  adjusted  to  it,  or  (still  better)  engraved  side  by 
side  upon  the  tube,  we  easily  see  how  to  reduce  from  one 
scale  to  the  other. 

/ 


F-H- 


12 


B!L 


80 


"  For  if/,  c,  r  be  the  various  readings  of  one  temperature, 

it  is  obvious  that 

/—  32  bears  the  same  ratio  to  (212  - 

that  c  bears  to 

and  r  bears  to 

/- J2  ^  J^  ^  r^» 
1 80         I  oo      80 


32,  or)  180 

100 

80 


"  Hence 


W 


124 


ARITHMETIC  OF  CHEMISTRY 


APPENDIX    D 

CALCULATION   OR   ATOMIC   WEIGHTS  OF  THE   MORE 

COMMON   ELEMENTS 


Name. 

Sym- 
bol. 

Weight. 

Name. 

Sym- 
bol. 

Weight. 

Aluminium  .     .     . 

Al 

27 

Iodine  .... 

I 

127 

Antimony 

3b 

120 

Iron .     .     . 

Fe 

56 

Arsenic   . 

As 

75 

Lead      .     . 

Pb 

207 

Barium    . 

Ba 

m 

Magnesium 

Mg 

24.5 

Bismuth  . 

Bi 

208 

Manganese 

Mn 

55 

Boron 

Bo 

II 

Mercury     . 

Hg 

2CX) 

Bromine  . 

Br 

80 

Nickel  . 

Ni 

58.5 

Cadmium 

Cd 

1 1 2.5 

Nitrogen 

N 

14 

Calcium  . 

Ca 

40 

Oxygen 

0 

16 

Carbon '  . 

C 

12 

Phosphorus 

P 

31 

Chlorine  . 

CI 

35-5 

Potassium 

K 

39 

Chromium 

Cr 

52 

Silicon  . 

Si 

28.5 

Cobalt     .     . 

Co 

59 

Silver     . 

Ag 

108 

Copper    . 

Cu 

63.5 

Sodium . 

Na 

23 

Fluorine .     . 

F 

19 

Strontium 

Sr 

87.5 

Gold  .     . 

Au 

197-5 

Sulphur 

S 

32 

Hydrogen    . 

H 

I 

Tin    .     . 

Sn 

119 

Zinc .     . 

Zn 

65.5 

I 


APPENDICES 


125 


127 

56 
207 

24.5 
55 

2CX3 

58.5 

14 

16 

39 

28.5 

108 

23 

87-5 

32 
119 

65-5 


APPENDIX   E 

EQUATIONS  IN  FREQUExNT  USE 

The  following  equations  are  among  the  most  commonly 
met.  Though  equations  are  made  to  represent  the  facts  as 
found  out  by  experiment,  the  beginner  often  finds  it  easiest 
to  remember  the  equation,  and  to  translate  it  into  a  verbal 
expression  of  the  experimental  fact.  In  the  following  equa- 
tions ^^j^j  are  represented  by  the  italics,  and  of  these  volumes 
may  be  calculated. 

OxVGliN 


;  / 


2HgO 

Mercuric 
oxide 

=      2  Hg  4-  0. 

3  MnO, 

Manganese 
peroxide 
(or  black  oxide  of  manganese) 

=          MngO  J  -f  O^ 

Manganoso- 
manganic  oxide 
(or  red  oxide  of  manganese) 

2  KCIO3 

=       2  KCl  +  3  C>2 

c  +  a 

-      CO. 

s-f  0. 

=     SO, 

Sulphur 
dioxide 

P4  +  5  0, 

=        2  P,05 

Phosphorus 
pentoxide 

3  Fe  +  2  a. 

=     Fe,0. 

Ferroso- ferric 
oxide 

126 


ARITHMETIC  OF  CHEMISTRY 


2KMn04  +  3H2SO4  4-  5H2O2 
Potassium  Hydrogen 

permanganate  peroxide 

:    K2SO4    +    2  MnS04    +    8  H2O    +    5  (?2 

Potassium       Manganese 
sulphate  sulphate 


1 4 

t 


Hydrogen 


2H2O 

2  K  +  2  H2O 

3  Fe  +  4  H2O 
Zn  +  2  HCl 
Zn  +  H2SO4 

ZnO  +  H2SO4 
CuO  4-  H2 


=     2  KOH  4-  Hi 

Potassium 
hydrate 

=  Fe304  4-  4  ^2 

=  ZnCl2  4-  Hi 

=  ZnS04  4-  Hi 

=  ZnS04  4-  H2O 

=  Cu  4-  H2O 


Nitrogen 


KNO2  4-  NH4CI 

Potassium  Ammonium 
nitrite        chloride 


=     KCl  4-  2  H2O  -h  Ni 


2  NH4CI 4-  CaO     = 
Lime 

CaClg  4-  H2O  4-  2NH^ 

Ammonia 

NH^  +  HCl    = 

NH4CI 

/^NH,  +  zO,     = 

2iV^2  4-6H20 

NH4NO,     = 

Ammonium 
nitrate 

2H20  4-^2<^ 

Nitrous 
oxide 

APPENDICES 


127 


3CU  +  8HNO3     = 


2Pb(N03)2       = 


3Cu(NO,3)2+4H,0+2iV(9 
Cupric  Nitric 

nitrate  oxide 

2PbO-f  0,^2N.,0: 

Nitrogen 
peroxide 


KNO3  +  H2SO4      =      KHSO4  +  HNO3 

Potassium 
acid  sulphate 

Carbon 

CaCOa     =     CaO  +  CO. 

CaCOg  +  2  HCl     =     CaClg  +  HoO  +  CO, 

C2H2O4     =     H2O  ^-C0-^  CO, 

Oxalic 
acid 

K4FeC6N6  +  6  H2O  +  6  H2SO4 
Potassium 
ferrocyanide 

=  2  K2SO4  4-  3(NH4)2S04  -f-  FeS04  +  6  CO 
2C0-\-  O.    =     2  CO. 
CO,  -h  C     =     2  CO 
4  Ca  +  3  Fe     =     Fe304  ^  ^CO 

NaCgHgOa  +  NaOH     =     NasCO,  +  CH, 

Sodium  Methane 

acetate  or  marsh  gas 


CH^  -V  2  O, 


CO.  +  2  H2O 


*  At  about  icx)°  C.  the  formula  of  nitrogen  peroxide  is  NO2. 


128 


AHITHMETIC  OF  CHEMISTRY 


Calcium 
carbide 

2  Co//,  +  5  O., 


Ca(OH),+  Q//, 
Slaked        Acetylene 
lime 

4  ca  -f  2  H2O 

C6>a+2^, 


Chlorine 

NaCl  4-  H,,SO^  =     NaHSO^  +  HCl 

2  NaCl  +  HoSO,  =     Na2S04  +  2  //C/ 

NaOH  4-  HCl  =     NaCl  +  H^O 

MnO,  4-  4  kci  =     MnCl,  +  2  H,0  +  Cl^ 

2CL  +  2  H,,0  =     4  HCl  +  a. 

,Cr.D,  +  14  HCl     =     2  KCH-2  CrCl3+  7  H,0+3  C/g 
»tassium  Chromic 


K 

Potassium 
bichromate 


Chromic 
chloride 


2  KOH  +  CI,     =     KCI  4-  KCIO 

Potassium 
hypochlorite 

6KOH  +  3C/0      =      5KCI4-KCIO3 
KCIO +  2  HCl     =     KCI  4- H2O  4- C/a 

3  KCIO3  4- 2  H.,SO,     =     KCIO44-2KHSO44-H2O4-CIA 

Potassium  Chlorine 

.....  peroxide 


Potassium 
perchlorate 


SJVB,  +  ^C/,     =     6NH4Cl4-iV2 
CoMa  4-  Cl>     =     C2H4CI2 

Ethylene  Ethylene 

dichloride 


t 
M 


APPENDICES 


129 


2>Ck 


orine 
oxide 


•  Iodine 


MnO,  -}-  2  KI  -f  2  H,SO.  =  K,SO^  +  MiiSO,+  2  H,0  +  I, 

2  KI  +  67,  =  2  KCl  -f  I, 

Fluorfnr 

CaF2-t-H,,S0,  =  CaS0,  +  2///?' 

SiO,  +  4y///  =  SiF,^-2Yi\0 

Silica  Silict)n 

lluoridc 


3  SiF,  4-  4  HoO 


H4SiO,  +  2H,,SiF„ 

Silicic         Ilydrolluu- 
acid  silicic  acid 


Sulphur 


FeSs     =     FeS  +  S 
Pyrite  herrous 

sulphide 

FeS  +  HoSO,     =     FeSO.  +  JhS 

Sb,S,  -f-  6  HCl     =     2  SbClg  +  3  iY,5 
Antimony 
trisulphide  ' 

2^2^+3  a       =        2HoO  +  26'6>2 

CUSO4  -j-  HS     =     CuS  +  H,S04 
HS  +  4  Bro  +  4  H2O     =     8  HBr  +  H2SO4 
Zf^o^+Ia     =     2HI-fS 

Phosphorus 

P4  +  3  KOH  4-  3  H,0     =     3  KH0PO2  +  PH, 

I'otassium     I'hosphine 
hypophosphite 

K 


, 


130 


ARITHMETIC  OF  CHEMISTRY 


2/^3  +  4  <^2     =     P2O5  +  3H2O 


2  H3PO2     = 
Hypophosphorous 
acid 


2  H3PO4  +  /W3 

Phosphoric 
acid 


4H3PO3     =     3H3P04  +  /'^3 

Phosphorous 
acid 


Ca3(P04)2+2H2S04 

Tricalcic 
phosphate 

CaH4(P04)2 


=     CaH4(P04)2  +  2CaS04 

Calcium  acid 
phosphate 

=     Ca(P03)2+2H20 
Calcium 
metaphosphaf.e 


3  Ca(P03)2  +  10  C     =     Ca8(P04)2  +  10  CO  +  P4 


1 


APPENDICES 


131 


APPENDIX   F 

VAPOUR  PRESSURE   OF  WATER  BETWEEN  THE 
TEMPERATURES  0°  C.  AND    100°  C. 

(taken  from  dittmar's  chemical  arithmetic) 


P4 


Temp. 

Vapour 
Pressure 

Temp. 

Vapour 
Pressure. 

Temp. 

Vapour 
Pressure. 

Temp. 

Vapour 
Pressure. 

0° 

4-57 

I 

4.91 

26 

24.96 

5» 

96.66 

76 

301.09 

2 

5-27 

27 

26.47 

52 

tOl.55 

77 

313.85 

3 

5.66 

28 

28.07 

53 

106.65 

78 

32705 

4 

6.07 

29 

29.74 

54 

III.97 

79 

340.73 

5 

6.51 

30 

31.51 

55 

117.52 

80 

354.87 

6 

6.97 

31 

33.37 

56 

123.29 

81 

369.51 

7 

747 

32 

35.32 

57 

129.31 

82 

384.64 

8 

7.99 

33 

37-37 

58 

135.58 

83 

400.29 

9 

8.55 

34 

39.52 

59 

142.10 

84 

416.47 

10 

9.14 

35 

41.78 

60 

148.88 

85 

433.19 

II 

9.77 

36 

44.16 

61 

155.95 

86 

450.47 

12 

10.43 

37 

46.65 

62 

163.29 

87 

468.32 

13 

II. 14 

38 

49.26 

63 

170.82 

88 

486.76 

14 

11.88 

39 

52.00 

64 

178.86 

89 

505.81 

15 

12.67 

40 

54.87 

65 

187.10 

90 

525.47 

16 

13.51 

41 

57.87 

66 

195.67 

91 

545-77 

17 

14.40 

42 

61.02 

67 

204.56 

92 

566.71 

18 

15.33 

43 

64.31 

68 

213.79 

93 

588.33 

19 

16.32 

44 

67.76 

69 

223.37 

94 

610.64 

20 

17.36 

45 

71.36 

70 

233.31 

95 

633.66 

21 

18.47 

46 

75.13 

71 

243.62 

96 

657.40 

22 

19.63 

47 

79.07 

72 

254.30 

97 

681.88 

23 

20.86 

48 

83.19 

73 

265.38 

98 

707.13 

24 

22.15 

49 

87.49 

74 

276.87 

99 

733.16 

25 

23.52 

50 

91.98 

75 

288.76 

100 

760.00 

\ 


132 


ARITHMETIC  OF  CHEMISTRY 


APPENDIX   G 

TO  FIND  THE  LOGARnilMS  OK  A  NUMBER 

Rule.  —  Regard  the  number  ()  as  separated  into  tivo 
factors  i\  X  10",  luherc  i\  begins  in  the  units'  phice.  Find 
in  the  tables  the  logarithm  of  (].  This  ivill  be  the  mantissa 
of  the  desired  logarithm.  Prefix  to  this  the  characteristic  or 
index  ±  n. 

A  few  examples  will  sufficiently  elucidate  the  process. 

Example.  —  Desired  the  logarithm  to  four  decimal  places  of  the 
number  306. 

Write  the  number,  or,  better,  merely  consider  it  as  if  factored  in  the 
form  3.o6'io-.  In  the  four-place  table,  on  the  hue  3.0  and  in  the 
column  headed  6  will  be  found  .4857,  which  is  log  3.06.  Obviously, 
log  10-  =  2.  Therefore  log  306.  =  log  3.06  +  log  10'^  =  4857  +  2, 
which  is  usually  written  2.4857. 

Further  Examples.  Interpolation.  —  Desired  to  four  decimal 
places  the  logarithm  of  306.2. 

This  will  lie  between  the  logs  of  306  and  307  (and  approximately 
0.2  of  the  way),  and  as  the  table  is  not  carried  out  further  we  must 
interpolate. 


For  3.07  we  find  4871 

For  3.06  we  find  4857 

Interpolation,  0.2   of    14 

=  2.8  =         3 

.'.  log  3.062  =  .4860 
.*.  log  306.2  =  log  3.062  -f  log  10'^ 
=  2.4860 


Difference  =  .0014,  usually  writ- 
ten 14. 

The  desired  mantissa  will  be  0.2 
of  the  way  from  .4857  to  .4871. 
Hence  we  murt  add  to  the  former 
number  0.214.  =  3. 


The  interpolation  may  always  be  made  by  subtracting  and 
multiplying  as  in  this  example,  but  to  save  the  labour,  log- 


» 


APPENDICES 


133 


aritlim  tables  are  usually  provided  with  marginal  interpola- 
tion tables,  ])y  the  aid  of  which  interpolations  may  easily 
be  made  mentally. 

Thus  in  taking  out  log  3.062  we  find  log  3.06  =  4857.  I5y  inspec- 
tion, tlilTercnce  =  14.  In  the  interpolation  table  headed  14,  line  2, 
stands  3,  which  is  therefore  the  desired  0.2  of  14.  Thereftirc  log  3.062 
=  .4860.     This  operation  could,  of  course,  i)c  carried  out  mentally. 

The  present  tables  are  arranged  so  that  the  interpolation  tables  stand 
opposite,  or  nearly  so,  to  the  logarithms  to  which  they  correspond. 
This  not  only  gives  them  a  convenient  location,  l)ut  enables  the  com- 
puter usually  to  avoid  even  the  mental  subtraction  of  the  successive 
logarithms  to  find  the  difference,  since  this  will,  of  course,  be  that  at 
the  head  of  the  nearest  difference  table.  Usually  also  the  error  intro- 
duced by  using  an  interpolation  table  slightly  too  large  or  too  small  is 
negligible. 


I 


C/) 

X 
H 

< 

o 
o 


III 

o 

< 

-J 

Q. 

D 
O 


5s 

£9    rovo 

0 

1^ 

vO    On  N  NO    On 
1-    i-i    <S    N    N 

S 

N    u^ 

t^  0 

^4 

»i4 

t^  ON  N 

N»     w     M 

»    N    ro  lovo 

jg  ror. 

0 

N4 

t~^  O    -t  «^  "- 
w    M    N    N    ro 

S 

ro  u^oo    0 

rONO 

00    *-    ro 
«     M     N 

00   (^   ^  m  t^ 

s  -^^ 

00    N    m  On   « 
i-i    N    N    cs    ro 

S 

rovO 

00      M 

""t 

N    N     N 

©    N    ''f  NO  00 

CO 

On  ro  r^  O     -f 
«    M    N    ro   fO 

S 

fOO 

On  N 

CI    N    N 

g    N    ■'f  rx  On 

a 

ro  ao    M  VO 
8.    8    0    0 

O 

On  0    0    0    0 
»t   ON   ro   1^  « 
M    n     ro  ro   Tf 
q    O    0    0    O 

ro 

q 

u^nO 

o  « 

O     M 

ro  ro 

On 
r» 

On  »r>  On 
«S   NO   00 
lO    t^    ON 

N   N   N 

►;    Tf  QO    Tf  M 
O    O    OnOO  no 
N    Tf  lo  r>.  ov 

to  ro  ro  ro  ro 

00 

8  8  0 

O 

o 

i/^  nO  nO  vO  NO 
TfOO    NO    O 
N    M    ro  ro   »t 

q  0  0  o  o 

ro 

On  N 
—    r>. 

0    '-' 

On  ro 
ON    0 
ro  t>i 

00 

On 

• 

ro 

LO 

N 

O     Tj-vo 
lO   1^   ON 
N     N     N 

«    u^  Ov  NO    VO 
00  00    I^nO    t 
1-1    ro  VO  1^  On 
ro  ro  ro  ro  ro 

t^ 

8  8  o 

00 
0 

0 

i-i    N    N    N    N 

TfOO     N   VO     O 
M     PJ     ro  ro   Tf 

q  o  0  o  o 

M-  N  0^    »^  ro 
On  00    ro  NO    t^ 

M   NO    O    ro  NO 

q  0  ■-•  1^  >^ 

On 
lO 

ON 

• 

M 

Q  00    "^ 

00       «       Tf 
Tf     t~^    ON 

M     N     N 

Q    "^0    l^  t^ 

NO  NO  NO   'f  n 

«    ro  VO  t^  Ov 
ro  ro  ro  ro  ro 

VO 

vO    On  w 

8.8  o 

in 

0 

in 

On 

O 

t^OO  00  00  00 
ro  i~-,  «    to  ON 

N     N    ro  ro  ro 

q  o  o  0  0 

ro 

lO^ 

rOvO 

ro 

ON 

o 

xo  lO  ro 
to  On  N 

fNO      ON 

W    M    N 

On  vn  w   On  0\ 
ro  Tf  •*  "    0 
I-I    ro  VO  t^  CN 
ro  ro  ro  ro  ro 

in 

Q    Q    w 

q  5  o 

5i 

0 

M 
ON 

0 

ro  ro  •<*•  ^  Th 
ro  t^  ►-    lo  ON 
M    P)    ro  ro  ro 
q    O    O    O    O 

ts.   ON 

Q  NO 

o   « 

rOvO 

M      11 

■ 

IT) 

O    N    Q 

ro  t^  0 

TfNO      ON 
N       N       CN| 

00    Tt  N    iM   N 

I-I     N     N    M    On 

►-    ro  m  t^oo 
ro  ro  ^  ro  ro 

-* 

8  8  5 

•^00 

o  o 

•iS  vg^  2    Po  §N 

S  S  S^  o'  o' 

q  ^8^ 

M     i-i 

u^OO 
t^  Tt 

00   « 

• 

u^OO  00 

O       Tf     t^ 

"^No  00 

N    W    N 

vO    Tf  N    N    Tf 

ON   O     O    On  t^ 
O    ro  voNO  00 

to  r^)  fO  to  to 

• 

«*> 

fO  NO     On 
M    "^  On 

?88 

0   o 

Tf  ionO  no  nO 
N  vO    O    '^00 
N    N    ro  ro  ro 
q    0    O    O    O 

00 
N 

M 

►-    On 
ro  On 
ir>00 
O    O 

On  ro 
ro  v.-> 

O    u^nO 

CO      N      IT) 

rONO   00 
N    N    N 

vr>  Tf  ro  Tf  VO 
t».oo  00   t>.  m 

O     N     ■'f  NO  00 

ro  ro  rO  ro  ro 

M 

On  N    lO 

0    "^  On 

88  8 

O 

ON 

0 

O    -    N    N    N 

N  vO    o    "i-00 
M    PJ    ro  ro  ro 
q    O    O    O    O 

VO 

• 

N    TfNO    ro 
ONNO     O     N 
rt  00    N    ir> 
0    0    "-•    «-" 

00 
00 

• 

lO 

to  M    ro 
lo  0    ro 
rOvO   00 
M     N     N 

Tf  ro  Tf  loOO 
vONO   NO    VO  ro 
O     M     "<f  NO  00 
ro  ro  ro  ro  ro 

• 

H 

^«5    0 

0 

IT) 

M 

o 

NO    r^OO  00  00 
M     lo  On  ro  r^ 
C)    rj    N    ro  ro 

q  0  0  o  o 

fO 

1 

rooo 

lO   N 

Z°8 

ro  N 
t^  On 

O  00 

On  NO 

t^  o 
I-;    C^ 

O     t^  O 
ro   t^  ii 
ro    U^OO 
N     N     N 

N    ro  Tf  NO    0 
ro  Tf  T-  ro  N 
O    N    4nO  00 
ro  ro  ro  ro  ro 

o 

O    <^nO 

• 

00 
N 

0 

0 

N    ro  "*  rf  Tj- 
>-    lO  On  ro  t^ 
N    N    N    ro  ro 
q    0    0    0    0 

1 

o  o 

On  « 
rOvO 

N4 

NO 

Tf  rooo 
O    LOOO 
ro  lo  i^ 
w    N    N 

O    N     Tf  t^  N 

O     N     ^vS  00 

CO    ro  ro  ro  ro 

• 

o 

Z 

pq  q 

? 

? 

ionO    t^OO    On 

q  q  q  q  q 

O 

• 

i-i    N 

•        • 

rOTj- 

u^vO 

• 

t^OO    On 

•            •            • 

p  «    «    ro  t 

00    O   •-   ro 

M      M      M 

ON  I-    ro  ^  ' 

M      M       M 

0     N     'f  NO   C 

M       M       M      M 

M    fO  vrjOO 

m^     ^m     tm     ^^ 

fO  C/5  NO    On 
ro  On  LO  Q 

«    r<    Tf  vo 

Tf     Tf     ^     Tf 

VO    -    0    'l- 

1-   00     Tf   ON 

I-I      M      Tf    LO 
Tf     Tf     4     ^ 

On  VO  lO  Ov  0 
On  O    M    r>i 
O    ri    Tf  lo 

Tf     Tf     4    Tf 

«    On  On  Tf  ( 

00    Tf  0  NO    » 

Q      P<      Tf    LO    1 
^     Tf    4     Tf 

ir>  N    rooo  0 
vO     ro  On   Tf   < 
0     f^     ro  LONj 
Tf    Tf    Tf    Tf    ' 

CO  vO  00    ro  r 
Tf  -    1^  rood 
O    PJ    ro  »ONi 

"^    t    Tf    Tf    , 

w    0    N  00    < 
ro  5   NO    -  N< 
O    N    ro  I'NNi 

^    Tf    Tf    Tf    1 

^    rONO    N    ' 
r-l  00    Tf   o    V 
O    -    ro  vrjNi 

^       Tf     ^     Tf     . 

r«*NO    O    t^  c 
ONNO    rooo    f 
On   «-    ro   ^  Nj 
ro  Tf  Tf  4  ' 

ON  O    rf  N    < 
t^  vT)  M    r^  ( 

On  «    ro  Tf  vj 
ro  Tf  Tf  4  " 

mvo   t^oo    < 

•  •  •  • 

r^ 


<0  N  ro  ir>\0 

00  N  «t  m  tx 

©  N  ^^O  00 

58  ^  ^  *^  c^ 


S  ^00  ^  M 
O  O  OnOO  ^ 
N  ^  «r>  t>.  0\ 
fO  fl  f)  ro  fO 


00  oo  r^o  "t 
M  ro  to  1^  a\ 
rn  fO  to  ro  ro 


Q  "^  O  !>.  r^ 

>0  vo  vO  »!■  N 

1-  ro  ir^  r^  Ov 

rn  rn  fO  m  r«^ 

OS  tr>  «    0\  0\ 

ro  •<t  '*  M  O 

I-  r^  u->  »>.  C\ 

r<^  to  fJ  O  fO 


00  ^  N  M  N 
i-i  N  N  M  0\ 
I-     fO  Ul  t>iOO 

fo  to  to  fo  fn 


vO  Tf  N    N    Tf 

ON  O  O    Cn  t^ 

O  fO  lOVO  00 

ro  ro  fO  ro  fO 


IT)  ^  ro  ThvO 
t>»QO  oo  t>.  m 
O  tJ  •^vO  00 
fO  CO  f)  fO  fO 


rf  fO  T^  iriOO 
UTsO  vO  "^  "^ 
O     N     -^O  00 

ro  ro  ro  fO  fO 


N  fO  TfvO  0 
ro  rh  '^  ro  N 
O  N  n-vO  00 
ro  f)  fO  fO  fO 


O  N  «t  t^  N 

i-t  N  N    «    Q 

O  N  't^  CO 

CO  fO  fO  ro  r<i 


i-i    N    fO  * 


00    O  **   ro  ^ 

M  M      >■      »4 

0\  11  fO  Tf  vO 

>«  M       »4       M 

O    N  '♦•"^  00 


■a     *■     M     1-4     C4 


M  M  C4   ^  tn  vO   t^oo  O  M 

w  Ml  fo  ^  tn  t>*oo  o\  O  N 

^  M  ro  TfvO  t^OO    O  «  «0 

V^  »4  N4  M 

to  M  ro  lr>^0  00    0\  1  N  "t 


00  M  N  N    rO  ^  tovo  vO    t<» 

O)  M  N  rO  ^  u^  w^vO    t^QO 

e  M  M  ro  '<t  mvO    t>iOO    OS 

iH  i-i  N  <0  rj-  \0    t^OO    OS  O 


ro  V5  vO    Cs  t^ 
ro  C3S  u^  O    in 

^  'if  4  "^  "* 


vO    «    O    't  N 

«   00    rf   ON   't 

Tf  rt   4  Tt  -^t 


OS  m  to  0\00 
ON  O     M     t^   N 

rt    Tl-    4   Tf    ^ 


I 


00     N     N   00 
fO  r^  O    M 

•»f  ir>  m  m  m 


0^   OS   ^   fO 
Tt   O   SO    w 


oo 

O      N      Tf 


ir>  N  rooO  00 
SO  ro  OS  Tf  OS 
O  M  ro  iO>0 
Tf   "^  "*  ^  ^ 


00  sO  00  PO  ro 
^  -  1^  rooo 
O  M  ro  mvO 
^  "t  •*  "*  ^ 


so  ^  0\  0\sO 

00  PJ    voQO    « 

QO  O    •-.    N    Ti- 

Tf  lo  m  »r>  irt 


n  f-l    u-ivO    ro 
t^tH    Ti-  i^  O 

00    ©     w     N     rf- 

^  lO  lo  to  io 

lO  On  fOsO    0\ 

00    OS  "-•    n    fO 

^  ^  ir>  ir>  VO 

fO  fO  OS  O  OO 
rf  00  «  lO  t>» 
00  OS  w  N  ro 
^  ^  to  to  >o 


rO   O 


M  OO    0\ 
_    vO    *<   SO 
O    P*    fO  lOvO 
"«1-   ">4-  rf   Tf   Tf 


1-t  00    ^  O    to 
O    "-    fO  lOvO 


OS  0\  to  tx.vO 

N  \0  O    fOsO 

00  OS  "I    N    fO 

^  5^  trt  tr>  to 

■»t  to  N    Th  fO 

w  irj  OS  N    tr> 

00  OS  O    N    fO 

T^  ^  tr>  to  to 


N    OS  w    O 

OS  O    M    ro 
Tt"  •^  to  to  to 


w  O  sO  On© 
in  t>.00  On  "H 
lOvO  t^OO  O 
lit  u^  I/)  tn  (O 


OsOO  ti^oo  OS 
ro  lO  t>.0O  On 
lOsO  t>.0O  OS 
\r\  LO  to  to  t/> 


tN.  r>i  rO  t^OO 

N  Tf  vO  t^QO 
too  t>.00  On 
IT)  to  tr>  to  to 


to  lO  N  so    t^ 
M   fo  losO   r^ 

to  so     t>iOO     OS 
ir»  to  tr>  to  to 


N  ro  O  tosO 
O  N  ^  to  SO 
vrjso  txOO  On 
in  to  to  tr>  to 


aM     OS   fO  to 
w     N     ^   to 

TfvO    t>.0O    On 
ir>  lO  to  to  lO 


00    On  »^  N    -^ 

t^  OS   ►-•     fO    'T 

Tf  to  t^OO    On 
lO  to  lO  to  to 


m  t^  to  M   ro 

so  00    O    M    fO 

^i-  lo  t^oo  0\ 

ir>  to  to  to  to 


11     N     N     PI     N 
w     P)     ro    -f   lO 

nO   nO   \0   nO   O 


r^  PI    »f  to  CO 

Q        NM       Nrt       »4       »^ 

►-     P4     'O    *<■    lO 

SO  sO  SO  SO  sO 


vO    M  tJ-   lO  ro 

OS  O  O    O    O 

0    P<  fo  "I-  to 

NO   nO  sO   sO   nO 


U1    M       Tf     lO    CO 

00  Ov  0\  OS  o» 
Q  «  P»  CO  -f 
SO  NO  sO  sO  NO 


to  Q  't-  lO  -f 

r>iO0  00  00  00 

Q      n  PI      CO    Tt 

SO    NO  SO    *0    NO 


oo  PI    cc^  CO  1 

►N  w    O    OS  OO 

NO  t^CC  CO    OS 

so  NO  nO  nO  nO 


$ 


PI       Tf     Tf     N 

O     On  00     t>. 
t^  l>.00    On 


SO   NO   >0   ^0 


On  CO  lo  to  ^ 

Ov  onoo    r-«.vo 

iOnO     l>.Ob     OS 
SO  NO  so  nO  nO 


Q    'I"  nO  nO    to 
OS  QO     I^nO     lO 

iOnO     t-»0O 
SO   NO   SO  SO 


^ 


^ 


to    t>.    t^NO 

t^  SO  lO  ^ 
to  NO  «^  00  OS 
so    NO   NO   sO   NO 


VO    NO 


O    T^  to  rh  1    moo  00   t^ 

t~>it^I-^t^  t^vOiOffPO 

P<     COtJ-  10\0     t^QO     OS 

nO   sO   O  SO   NO  NO  NO  SO 


fOOccjin^     '-nOOnOsOO 
iOnOsOnOnO      sOiOTfcrjPI 

"      -  too    r^oo 

NO   vO   NO  NO 


Q     ».     CI     ro   'l- 

sO  sO  sO  sO  NO 


S 


PJO\coiO^  inOOnOO 

Tfrfmiou")  lOTfrOfON 

Q    n    N    CO   't  ionO    t^OO    OS 

NO  SO  so  so  so  nO  vO  sO  sO  sO 


09 


I^nO 

0 

l>N. 

OS 

SO 

00 

moo 

00 

ro 

lO 

•"t 

OS 

PI 

M 

00 

CO 

to 

"+ 

PI 

r^ 

0 

NM 

»4 

OS  NO 

CO  00 

CO 

PO 

PI 

so 

OS 

PI 

to 

t^ 

On 

(g 

PI 

ro 

CO 

Tf 

'+ 

■<1- 

■* 

CO 

CO 

PI 

N4 

OS 

N4 

CO 

5 

NO 

c* 

ON 

o 

»1 

CO 

rf 

lONO 

On 

^. 

N4 

PI 

CO 

"+ 

IONO 

t>.00 

s 

CO 

• 

-* 

-■I- 

't 

^ 

"* 

lO 

to 

to 

to 

• 

to 

to 

to 

to 

NO 

NO 

NO 

so 

SO 

• 

NO 

NO 

NO 

OS 

o 

rh 

N 

Th 

h^ 

"<t 

M 

lO 

to 

N4 

CO 

N 

00 

11 

M 

00 

PI 

to 

lO 

PI 

00 

,_ 

N 

N 

t^ 

lO 

N« 

r^ 

pj 

t^ 

N^ 

moo 

m^ 

5 

NO 

00 

ON 

N4 

PI 

PI 

CO 

CO 

CO 

CO 

PI 

PI 

1-4 

O 

OS 

CO 

^ 

NO 

t~^ 

On 

O 

Nil 

CO 

ionO 

t^ 

ON 

n8 

• 

N4 

PI 

CO 

rj- 

to  NO 

t^oo 

2^ 

to 

• 

■^ 

"* 

"«*• 

">t 

't 

to 

to 

to 

to 

• 

to 

lO 

to 

to 

NO 

NO 

NO 

NO 

NO 

• 

NO 

NO 

>o 

NO 

to  so 

• 

«^00 

^ 

o 

¥^ 

p» 

ro 

"* 

to  so 

t««00 

ov 

o 

M 

N 

"? 

t 

lOvO 

t^oo 

•          • 

<^ 

N 

^ 

CO 

PO 

^^ 

Tf 

^ 

^ 

en 

z 

H 

ttr 

< 
o 
o 

o 

< 

Q. 

D 
O 


z 
o 

§1 

Si 


00 


1A 


CO 


f« 


^MMCtM      r^TfTfiz-iu^       lOl-lM^4M 


^.wwMfO     T}-^inOvO         CO»<MHf< 


vft     ID  fO   ""     OS      r^   IT)  M     O     I'* 
O     «     rj     r^   ro      "i"    IT)  O     «^   t>» 


-f   "-  oo 
00     On 


OS   Q 


00 


On    -f    On    ft 

00     "^  —    C/3 

00  00  00  OO  t 


ON   fO'O    CO    '/D 

O     -     PI     ro   ro 
t^  t^  t>«.   t^  t-^ 


O  »^X  O    Q 

IT)  rr)   «.  O    Cio 

O  «     N  '■O   r<^ 

t-^  t^  r^  f*  t^ 


N  -O  O    ft  N 

tJ-  P4  ^    ON  r^ 

O     «  N     CI  PO 

t^  t^  t>»  t>.  t>. 


POOO  N  "t  -t 
PO  «-"  O  00  vO 
O  "-I  N  M  PO 
t^  t>i  t>.  t^  t>« 


rl  1^  ON  t>«  lo 

O  •-''-<  N  PO 
t>.  t».  t>.  t-x  t>. 


vO  «  m  t^OO 

«  O  00  vO  "i- 
O  -<  w  N  PO 


o  po  o>  ^  r>» 

O  Tf  "«  Os  vO 

»^  in  o  so  t>» 

t>»  f^.  Ix  t>.  !>. 


OvOOOOO  PICOMsOON 

'•n"-oo'+"-  co-t—   t^fo 

OOONOsO""  -•ftPOP'^-t 

tx  t>i  t>.oo  00  00  00  00  oo  00 


On  vO    N  so 
ir>  PO  «  00 
•^  m  >o  >o    r>. 
r-«.  t>.  t>.  t>.  t>. 


5 


MPOPO«OsO>-isOON 
00     ON   Os   O     «       "-•     PI     "-n   '•'5   -f 

t^  t-«.  t>.oo  00    00  00  00  00  00 


sO    !>•  t>»  so    in 

O    so     N   00     t 

in  in  so  so    t>. 

oo  00  00  00  oo 


8      1-1      M       M      ^ 
O     W   CO    '1 

in  1  n  so  sO    t>. 

oo   00  00  00  00 


»+•  in  in  «/" 

0\  in  «    t^. 

't  m  sO  sc 
00  00  00  o- 


w  00    -t  On  N 
in  N    O    t>.  in 
Tf  inso  so    r>» 

1^   t>»  1^  !>.  t-«. 


po  O  t^  N    »n 

Tf   PJ  On  t>.  ^ 

^  in  inso    I--. 

t>.  t^  t^  t>.  t>i 


m  PP^  On  Tt- 00 
rn  «  00  >o    PO 
rf  in  in  so    t^ 
t^  t>.  t^  t>.  t>. 


t>.  in  N    t^  <-i 

M    O  oo   in  po 
rf-   i^  lOsO    t^ 


m  sO  so    in  N  On  1*^  On  'O  O 

PI     ON  so     rO   O  SO     PI   On  O     PJ 

QOOOOnO"^  11    pi    pi    pn^ 

tx  tx  t-^oo  00  00  00  00  00  00 


00  On 
CO  "t 
•H-  in  _ 

00  00  00 


f; 


fo  po 


m  ^ 


00  ON 

■->  00 

00  00 


On  00  so  PI  00  PO 

in  N  ON  so  PI  o^  m  N 

On  O  O  w  PI  PI  PI  't 

»xOO  CO  00  00  00  00  00 


rx  O 


CO 


PO  PO  P')  P, 
^   Q   so    PI 
Tf  mso  o    t>. 
00  00  00  00  ;o 


O    N    N    «    On 

wQO    in  PI  CO  -...-^    -.-^ 

OOOOOnQQ  wPIMPOtI- 

tx  tx  txoo  CO  oo  00  CO  00  00 


vO    PJ    rx  w    "il- 
in  N  00   I'N  « 


vO  tx  i^  r^O 
t"'*  pn  On  in  "" 
Tf  in  in  o    t>i 

CO  00  00  CO  jo 


po  m  m  ^  PI 

O  .«>  ^  «  CO 


On  iTN  0    ^  t^ 

Tf    «   00     1-    O 
OOOnOO      "iPJPIPn^ 

t-^  txco  CO    00  00  00  CO  00 


t-   pn  tx  On  o 

§on  !>«.  m  Tf 
O     ►-"     N     PO 
t«     t»  tx  t>.   tx 


OO  ^00  11  N 
On  00  sO  IT)  PO 
On  O    «    IN    PO 

SO  tx  tx  tx  tx 


On  rx  ^  On  PO 
w  On  tx  Tf  PI 
Tf  Tf  inso  tx 
t-s.  t-N.  t->  r>.  t>. 


so  00  00  rx  m 

ON  so  PO  O  •'^ 
t^OO  On 
tx  l>»  tx 


On  Tf  00  w 

,  O  l~«.  PO  O 

Q   w  N  N  PO  Tf 

CO  00  00  00  00  00 


"I- 


N  Q  SO 

n   ON  ' 


_   PJ  sO 

_   SO   Tf  M 

Tf  Tf  mso  tx 
tx  r^  tx  ix  1^ 


ONO'lOONSOPJt->«l1lA 

oosopopo     poosOpnoN 

t^OO     ON©    O       II     PI     M     PO  PO 

tx  t--.  txoo  CO    00  00  00  00  00 


t~>.  PO 
Tf  in 

00  00 


po  m 

so  N 

Tf  m 

00  00 


tx  On 
vn  w 

Tf  m 

00  00 


1  w  0 
On  in  1 
in  >0  i->» 

00  00  CO 


m  in  »* 

00 
I/- 


00 


o 
in' 

00 


KT  O 

00  00 


OnOO 
PO  ON 


On  I^^ 
sg^" 


PO  Tf 
-   Tf  N 

O  •*  «^  PO 


Tf  N  On  Tf  On 
O  20  in  PO  O 
Tf  Tf  inO  t-» 


pjpiTfpnPi      onu^mu^oo 

COit^PIOnO       PIONSOPIOO 

t^oo   ctnono     nwpipnpo 


:c2 


II  pp)  pn  po  PI 

m  M   t>i  po  ON 

Tf  m  in  so  so 

0000000000      ooooooooco 


I 


«    rt   p<^  Tf    insO    t^co    ON      O  >^    fi   m  ri-    mvo   t^oo   on 


w    N     PP)   ' 


tf» 


m 


;o 


vo 


• 

ON    l 

m  " 

00   < 

00  0 

1^  -^  c 

On  in  " 

00  CO  0( 

1 

> 

00    ' 

to  V. 

00 

• 

00  « 

On  t>.  P* 
tx  PO  0 
t^OO  oc 

00  00  00 

tx  POOO 

r>.oo  00 

oo  00  00 

00    lO  N 
vO    N  00 

IX 00   00 
00  00  00 

N    O  so 
vO    P<    l> 

txCO   00 

oo  00  00 

vo      Tf    W 

m  «    t> 

txCO   00 
00   00   00 

N4 

tx 

00 

00    "■ 

Q  so 

msO    b 

tx 

lO  >-i  "<  P<  N 


<0  M  M  r)  r( 


CO  C(0  00  00  CO 


Q       1^       N^       M 

O    O     N   00 

00   00   00    00 

• 

OS 
00 

-^   i/^  «r^  lr^ 

On  >0  "«    t^. 

00  00  00  c/w 

00    On  Cn 
00    "t  5  ^ 
•^-  in  '■o  >- 

00   00  00  0 

CI    ro  ro  r<^ 

00    Tf   Q   vo 

't  invo  o 

00  00  00  00 

p. 
00 

t>»  ro  On  lO  -< 
rf  >i->  >0  O  t>. 

00  00  00  00  00 


o 

M 

hM 

^N 

0 

r^ 

ro 

On 

lO 

*^ 

'f 

»o 

mvO 

t^ 

od 

00 

00 

00 

00 

m 

>n 

m 

U-) 

^ 

^ 

vO 

N 

00 

'4- 

.  ^ 

d 

Tf 

>ri 

IT 

■*ivn 

r^ 

00 

• 

00 

oc 

f^ 

00 

1^ 

ON 

o 

L  c^oo 

U-) 

N4 

t^ 

fro 

(T\ 

-"t 

m 

m 

l>o 

vO 

00 

00 

00 

f 

J 

M 

fO 

ro 

r^ 

M 

4 

m 

»4 

t^ 

fO 

c> 

pi 

T^ 

m 

mvO 

^ 

1 

00 

00 

00 

00 

oo 

»« 

w  N  ro  ' 


tn 

m 

Tt- 

^ 

m 

lO 

M 

M 

N 

N 

PO 

fO 

•* 

■♦ 

m 

^ 

0 

M 

M 

N 

N 

N 

PO 

m 

^ 

ro 

-«»■ 

•* 

m 

u^ 

« 

M 

M 

M 

PI 

PO 

"<!• 

'J- 

m 

m 

to 

t^ 

*^ 

PI 

PI 

PO 

fO  ^ 

^ 

m 

PI 

Os 

m 

«n 

On 

mo 

00 

<^ 

o 

a 

C, 

^ 

00 

NO 

m 

r^ 

r*^ 

QO 

PO 

CN| 

^. 

m 

to4 

t^ 

p« 

r>. 

rnoo 

m 

5 

On 

T  ■ 

On 

m 

00 

m 

Qn 

PI 

r>. 

M 

ON 

^  1 

m 

^^ 

3§ 

^ 

^ 

g. 

■ 

^4 

ON 

On 

On 

PI 

ON 

On 

^ 

in 

On 

mvo 
On  On 

• 

ON 

On 

IN. 

ON 

■§. 

^ 

^ 

» 

t^ 

•^ 

0 

in 

R 

'f'jO 

c8 

PI 

Th 

m 

m 

m 

^ 

m 

mt 

00 

m 

PI 

ss 

rt 

ON 

moo 

M 

On 

m 

»*4 

O 

1^ 

fS 

moo 

moo 

m 

m 

00 

PI 

r^ 

PI 

»4 

>n 

C 1 

Wf 

0\ 

r^oo 

5^ 

0\ 

o 

^ 

N4 

PJ 

PI 

m 

m 

"t 

Tf 

m 

mvo 

NO 

»^ 

r>. 

OO 

^ 

cs 

^ 

^K 

00 

00 

00  00    On 

ON 

ON 

On 

On 

On 

• 

On 

ON 

ON 

ON 

On 

ON 

ON 

On 

On 

q^ 

ON 

On 

ON 

'    ^ 

00 

^ 

5 

in 

CN 

N 

m 

t~>i 

ON 

0 

5 

0 

On  00 

NO 

rf 

»4 

I^ 

m 

cl 

Tf 

^ 

m 

t^ 

> 

Tt- 

1 

N4 

*o 

fl 

t-«. 

PI 

t-«. 

m 

m 

r«. 

PI 

t^ 

PI 

r>. 

M 

NO 

m 

^00      I 

»-«. 

po 

Qs 

o 

Q 

NN 

NN 

N 

P« 

m 

m 

t 

^ 

m 

inO 

NO 

1--. 

t^ 

00 

ON      I 

t 

KC 

o6    On 

^ 

ON 

ON 

On 

ON 

On 

• 

ON 

On 

ON 

ON 

On 

On 

On 

On 

ON 

• 

On 

ON 

ON 

^l    "^ 

N 

ON 

Tf 

5^ 

rn 

t>i 

o 

PJ 

■>t 

m 

in 

m 

't 

m 

M 

On  NO 

m 

On 

m 

o 

^ 

a\ 

PO     1 

L-?.^ 

Tf 

(2^ 

«n 

Q 

VO 

^N 

t>« 

P< 

ft. 

PI 

t-. 

PI 

r-«. 

PI 

r^ 

^m 

vO 

N4 

>n 

eg 

m 

^ 

moo    I 

oe 

On 

Q 

% 

^N 

NN 

P» 

PI 

f^ 

<-n 

■^ 

Tf 

in 

mNO 

NO 

t^ 

t>. 

00 

2> 

On    I 

00 

00 

00  00 

ON 

On 

On 

On 

On 

On 

ON 

On 

ON 

On 

On 

• 

On 

On 

ON 

ON 

<^ 

ON 

ON 

On  ON    1 

ON 

1^ 

On 

? 

S 

00 

m 

in 

NO 

S^ 

R 

O 
1^ 

^^ 

00 

N4 

^ 

NO 

=§ 

^ 

1 

m 

^ 

^ 

00 

1^00 

00 

Ov 

^5 

Sn 

• 

NN 

Mt 

PJ 

PI 

fo 

rO 

"^ 

T}- 

m 

mO 

NO 

t>. 

t>. 

On 

On 

00 

• 

00 

00 

00 

s 

o\ 

ON 

ON 

ON 

On 

• 

ON 

On 

On 

On 

<^ 

ON 

ON 

ON 

ON 

<^^ 

On 

ON 

ON 

ON 

T»- 

N4 

t^  rooo 

rovO 

On 

P» 

fO 

m 

m 

lO 

m 

p^ 

N 

^ 

r^ 

m 

o 

m 

N« 

NO 

O 

Tj- 

t>. 

r^OO 

T^ 

ON 

m 

o 

m 

^14 

vO 

M 

o 

N4 

NO 

M 

NO 

m 

o 

in 

On 

^ 

o8 

m 

r^ 

r^oo 

00 

On 

ON 

8n 

• 

N^ 

N^ 

P» 

PI 

m 

m 

■* 

Tj- 

in 

m 

NO 

t->i 

t^ 

1^00 

2» 

ON 

00 

• 

00 

00 

00 

00 

On 

On 

ON 

ON 

On 

• 

ON 

ON 

ON 

ON 

q^ 

ON 

ON 

ON 

On 

ON 

• 

On 

On 

On  ON    1 

s§ 

00 

00 
rn 

o\ 

t^ 

O 

m 

^ 

00 
in 

8^5 

0 

n8  S' 

m 

5" 

ON 

in 

On 

NO 

m 

^4 

VO 

^ 

t>»00 

00 

0\ 

ON 

• 

NN 

^N 

PJ 

PI 

m 

m 

^ 

■"t 

in 

m 

vO 

t^ 

I>nO0 

On 

8^ 

00 

• 

00 

00 

00 

00 

ON 

On 

ON 

ON 

On 

On 

ON 

ON 

On 

«^ 

c^ 

ON 

On 

On 

^ 

ON 

ON 

ON 

M 

o 

NO 

N 

t^ 

N 

NO 

On 

friN 

m 

T^ 

m 

IT 

m 

rf 

PI 

1 

r^ 

'^ 

M 

VO 

N 

t"* 

M 

m 

vO 

w 

!>. 

(^OO 

"* 

•<t 

o 

m 

o 

in 

O 

m 

O 

in 

'I- 

On 

"* 

00 

m 

t^ 

PI 

NO 

t^OO 

00 

ON 

^ 

8n 

• 

Q 

^^ 

N 

PI 

m 

fO 

■^ 

Tf 

in 

m 

NO 

NO 

r^ 

t^oo 

00 

OS 

ON 

00 

• 

00 

00 

00 

On 

On 

ON 

On 

On 

On 

ON 

ON 

ON 

On 

• 

ON 

On 

ON 

ON 

ON 

• 

On 

ON 

ON 

ON 

VO 

-t 

M 

r- 

N 

v£> 

§N 

f^NO 

00 

ON 

0 

8 

0 

On 

t^ 

m 

m 

ON  NO 

N 

t^ 

N 

tNi 

M 

xn 

N^ 

t>. 

N 

00 

fO 

rf 

On 

•<t 

CJN 

in 

in 

On 

Tf 

ON 

■*00 

m 

00 

PI 

t>. 

N^ 

NO 

t^oo 

00 

ON 

C3N 

§^ 

M 

NH 

M 

PI 

m 

-* 

•^ 

Tf 

m 

m^o 

NO 

r^ 

IN.  00 

00 

On 

ON 

00 

• 

00 

00 

00 

00 

ON 

c^ 

c^ 

0\ 

CN 

• 

On 

c^ 

On 

ON 

On 

On 

On 

<^ 

On 

<^ 

ON 

On 

On 

ON 

M 

00 

in 

ki4 

vO 

li^ 

in  00 

N4 

m 

T^ 

m 

m 

m 

Tf 

PI 

Sn 

00 

m 

M 

t^ 

moo 

N 

VO 

U-) 

c2 

vO 

N 

t->. 

fOOO 

ro 

ON 

rj- 

On 

Tf 

On 

ON 

Tj- 

moo 

m 

t^ 

N 

vO 

Ni^ 

m 

t^ 

00 

ON 

On 

Q 

o 

N^ 

PI 

PI 

m 

m 

^ 

m 

lOvO 

NO 

t>. 

r^oo 

00 

On 

ON 

00 

• 

00 

00 

00 

00 

On  On 

On 

On 

ON 

On 

On 

ON 

ON 

ON 

On 

ON 

ON 

ON 

On 

ON 

■ 

On 

ON 

ON 

ON 

•             • 

t^oo 

•        • 

On 

• 

P 

PJ 

fO 

1- 

tnvo 

•          • 

tN.00 

•          • 

On 

o 

• 

PI 

• 

m 

• 

t 

mNO 

•          • 

•          • 

J 

1 

OO 

00 

a> 

ON 

1 

THE  PRACTICAL  METHODS 


OF 


ORGANIC  CHEMISTRY 


AUTHORIZED  TRANSLATION 


lamo.    Cloth.    Price,  $1.60,  net 


It 


BY 


LUDWIG  GATTERMANN,  Ph.D., 

I'ro/essor  in  University  0/ Heidelberg. 


TRANSLATED  BY 

WILLIAM  SHAFER,  Ph.D., 

Instructor  in   Organic    Chemistry 
in  Lehigh  University. 


THE  GUARDIAN. 


"  The  selection  and  judjjmcnt  throughout  is  excellent.  The  book 
is  a  most  useful,  practical  adjunct  to  any  good  text-book  on  organic 
chemistry." 

PHARHACEUTICAL  REVIEW. 

"  This  is  a  book  that  should  be  in  the  library  of  every  teacher  of 
organic  chemistry,  and  one  which  wi''  no  doubt  be  of  great  value 
to  students  in  their  second  year  of  organic  chemistry.  Its  chief  pecu- 
liarity and  merit  is  in  the  great  stress  laid  on  practical  laboratory  work. 
...    It  is  permanently  a  worker's  guide." 

NATURE. 

"  Since  the  advance  of  organic  chemistry  in  this  country  must,  in  a 
measure,  depend  on  the  nature  of  the  availal)le  text-books,  both  the 
author  and  the  translator  deserve  our  thanks  for  providing  us  with 
a  work  such  as  the  present  one." 


PUBLISHED   BY 

THE    MACMILLAN    COMPANY 

66   FIFTH   AVENUE,    NEW  YORK 


I 


OUTLINES 

OF 

INDUSTRIAL  CHEMISTRY 

A  TEXT-BOOK   FOR  STUDENTS 
By   FRANK    HALL  THORP,   Ph.D., 

Instructor  in  Industrial   Chemistry  in   the  Massachusetts  Institute 

of  Technology. 

Cloth.    8vo.    Price,  $3.50  nei 


JAMES  LEWIS  HOWE. 

Department  of  Chemistry y  Washington  ami  Lee  University. 

"The  book  is  brought  thoroughly  up  to  date,  and  in  some  cases  the 
Hnes  of  probable  ilevelopmcnt  are  nicely  foreshadowed.  The  descrip- 
tions are  particularly  .lucid  and  the  illustrations  wdl  selected. 

The  general  arrangement  and  make-up  of  the  book  is  excellent,  and 
. .  .  altogether  the  book  tills  well  a  need  long  felt  by  teachers  of  Indus- 
trial Chemistry. 

I  shall  adopt  the  book  for  my  class  and  shall  take  pleasure  in 
recommending  it." 

CHARLES  E.  COATES.  Jr.,  Ph.D., 

Professor  of  Chemistry,  Louisiana  State  University. 

"  I  have  examined  it  carefully  and  think  it  a  most  excellent  book, 
meeting  a  want  I  have  long  felt  in  my  higher  classes.  I  have  intro- 
duced it  in  this  year's  classes." 

W.   A.   NOYES,  in   Science. 

"  The  descriptions  of  processes,  while  necessarily  concise,  are  clear 
and  interesting.  The  author  has  evidently  made  a  careful  study  of 
recent  methods  of  manufacture  as  well  as  of  older,  standard  processes. 
The  frei|uent  reference  to  American  practice  is  an  important  feature 
which  distinguishes  the  book  from  other  works  on  chemical  technology. 
A  select  bibliography  follows  each  subject,  and  will  be.  found  very 
useful." 


PUBLISHED   BY 

THE    MACMILLAN    COMPANY 

66   FIFTH   AVENUE,   NEW  YORK 


ent  book, 
ave  intro- 


are  clear 
study  of 

processes. 

lit  feature 

chnology. 

•and  very 


'RY 


\  ^ 


\ '- 


1/ 


stitute 


ir 


cases  the 
e  descrip- 

lUent,  and 
of  Indus- 

easure  in 


